虽然普及组一般都是暴力省一，但是有一些题目还是挺难的qwq个人觉得能进TG的题目会在前面打上'*'
NOIP2002(clear)
1 #include
2 using namespace std;
3 int main(){
4 std::ios::sync_with_stdio(false);
5 double Sn = 0;
6 int k , i = 1;
7 cin >> k;
8 while(Sn <= k) Sn += 1.0 / i++;
9 cout << --i;
10 return 0;
11 }
1 //递推
2 #include
3 using namespace std;
4 long long num[21][21];
5 int dir[8][2] = {1,2,1,-2,-1,2,-1,-2,2,1,2,-1,-2,-1,-2,1};
6 int main()
7 {
8 ios::sync_with_stdio(false);
9 int a , b , c , d;
10 cin >> a >> b >> c >> d;
11 for(int i = 0 ; i <= a ; i++)
12 for(int j = 0 ; j <= b ; j++)
13 {
14 int f = 1;
15 for(int k = 0 ; k < 8 && f ; k++)
16 if(c + dir[k][0] == i && d + dir[k][1] == j) f = 0;
17 if(f && (c != i || d != j))
18 if(i + j) num[i][j] = num[i ? i - 1 : 0][j] + num[i][j ? j - 1 : 0];
19 else num[i][j] = 1;
20 }
21 cout << num[a][b];
22 return 0;
23 }
1 //递归
2 #include
3 using namespace std;
4 int num[21] , cou , n;
5 bool vis[21];
6 bool ifZ(int a)
7 {
8 if(a == 1 || !a) return false;
9 for(int i = 2 ; i * i <= a ; i++)
10 if(a % i == 0) return false;
11 return true;
12 }
13 void dfs(int a , int b , int c)
14 {
15 if(!a)
16 {
17 if(ifZ(b)) cou++;
18 return;
19 }
20 for(int i = c ; i <= n - a ; i++)
21 if(!vis[i])
22 {
23 vis[i] = 1;
24 dfs(a - 1 , b + num[i] , i + 1);
25 vis[i] = 0;
26 }
27 }
28 int main()
29 {
30 std::ios::sync_with_stdio(false);
31 int k;
32 cin >> n >> k;
33 for(int i = 0 ; i < n ; i++) cin >> num[i];
34 dfs(k , 0 , 0);
35 cout << cou;
36 return 0;
37 }
1 //按位搜索
2 #include
3 using namespace std;
4 multimap < char , char > rule;
5 bool times[31][10];
6 int num[31] = {1 , 1};
7 int dfs(char a , int b)
8 {
9 if(times[b][a - '0']) return 0;
10 times[b][a - '0'] = 1;
11 int sum = 1;
12 map < char , char > :: iterator iter = rule.find(a);
13 for(int i = 0 ; i != rule.count(a) ; ++i , ++iter)
14 sum += dfs(iter -> second , b);
15 return sum;
16 }
17 int main()
18 {
19 string s;
20 int n;
21 for(cin >> s >> n ; n ; n--){
22 char a , b;
23 cin >> a >> b;
24 rule.insert(make_pair(a , b));
25 }
26 for(int i = 0 ; i < s.size() ; i++)
27 {
28 int k = dfs(s[i] , i);
29 for(int i = 1 ; i <= num[0] ; i++) num[i] *= k;
30 for(int i = 1 ; i <= num[0] ; i++)
31 if(num[i] >= 10)
32 {
33 num[i + 1] += num[i] / 10;
34 num[i] %= 10;
35 if(i == num[0]) num[0]++;
36 }
37 }
38 while(num[0]--) cout << num[num[0] + 1];
39 return 0;
40 }
NOIP2003(clear)
1 //卡特兰数
2 #include
3 #define ll long long
4 using namespace std;
5 inline ll fast_read()
6 {
7 char c = getchar();
8 ll a = 0;
9 int k = 0;
10 while(!isdigit(c))
11 {
12 if(c == '-') k = 1;
13 c = getchar();
14 }
15 while(isdigit(c)) a = (a << 3) + (a << 1) + c - '0' , c = getchar();
16 return k ? -a : a;
17 }
18 inline void fast_print(ll x)
19 {
20 int num[1001] , dirN = -1;
21 if(x < 0) putchar('-') , x = -x;
22 while(x) num[++dirN] = x % 10 , x /= 10;
23 if(dirN == -1) putchar('0');
24 while(dirN + 1) putchar(num[dirN--] + 48);
25 putchar('\n');
26 }
27 int num[19] = {1 , 1};
28 int main()
29 {
30 int n = fast_read();
31 for(int i = 2 ; i <= n ; i++)
32 for(int j = 0 ; j < n ; j++)
33 num[i] += num[j] * num[i - j - 1];
34 cout << num[n];
35 return 0;
36 }
1 //注意最后的0:0
2 #include
3 using namespace std;
4 int ans[2500][2];
5 int main()
6 {
7 string s;
8 int W11 = 0 , L11 = 0 , W21 = 0 , L21 = 0 , dirA = 0;
9 while(cin >> s)
10 {
11 int ifE = 0;
12 for(int i = 0 ; i < s.size() && !ifE ; i++)
13 {
14 if(s[i] == 'W')
15 {
16 W11++;
17 W21++;
18 if(W11 >= 11 && W11 - L11 >= 2)
19 {
20 cout << W11 << ":" << L11 << endl;
21 W11 = L11 = 0;
22 }
23 if(W21 >= 21 && W21 - L21 >= 2)
24 {
25 ans[dirA][0] = W21;
26 ans[dirA++][1] = L21;
27 W21 = L21 = 0;
28 }
29 }
30 else if(s[i] == 'L')
31 {
32 L11++;
33 L21++;
34 if(L11 >= 11 && L11 - W11 >= 2)
35 {
36 cout << W11 << ":" << L11 << endl;
37 W11 = L11 = 0;
38 }
39 if(L21 >= 21 && L21 - W21 >= 2)
40 {
41 ans[dirA][0] = W21;
42 ans[dirA++][1] = L21;
43 W21 = L21 = 0;
44 }
45 }
46 else ifE = 1;
47 }
48 if(ifE) break;
49 }
50 cout << W11 << ":" << L11 << endl;
51 for(int i = 0 ; i < dirA ; i++) cout << endl << ans[i][0] << ":" << ans[i][1];
52 cout << endl << W21 << ":" << L21;
53 return 0;
54 }
1 //区间DP
2 #include
3 using namespace std;
4 int num[101] , maxN[101][101][10] , minN[101][101][10];
5 inline int max(int a , int b){
6 return a > b ? a : b;
7 }
8 inline int min(int a , int b){
9 return a < b ? a : b;
10 }
11 int main(){
12 memset(minN , 0x3f , sizeof(minN));
13 int N , M;
14 cin >> N >> M;
15 for(int i = 1 ; i <= N ; i++)
16 {
17 cin >> num[i];
18 num[i + N] = num[i];
19 }
20 for(int i = 1 ; i <= N << 1 ; i++)
21 for(int j = i ; j <= N << 1 ; j++)
22 minN[i][j][1] = maxN[i][j][1]
23 = ((maxN[i][j - 1][1] + num[j]) % 10 + 10) % 10;
24 for(int k = 2 ; k <= M ; k++)
25 for(int i = 1 ; i <= (N << 1) - k + 1 ; i++)
26 for(int j = i + k - 1 ; j <= N << 1 && j - i + 1 <= N ; j++)
27 for(int m = i ; m < j ; m++)
28 for(int n = max(1 , k - j + m) ; n < k && n <= m - i + 1 ; n++)
29 {
30 maxN[i][j][k] = max(maxN[i][j][k] , maxN[i][m][n] * maxN[m + 1][j][k - n]);
31 minN[i][j][k] = min(minN[i][j][k] , minN[i][m][n] * minN[m + 1][j][k - n]);
32 }
33 int maxM = 0 , minM = 0x3f3f3f3f;
34 for(int i = 1 ; i <= N ; i++)
35 {
36 maxM = max(maxM , maxN[i][i + N - 1][M]);
37 minM = min(minM , minN[i][i + N - 1][M]);
38 }
39 cout << minM << endl << maxM;
40 return 0;
41 }
1 //第一问用对数函数得答案，第二问高精快速幂
2 #include
3 using namespace std;
4 int num1[502] = {0 , 2} , num2[502] , num3[502] = {0 , 1};
5 int main()
6 {
7 std::ios::sync_with_stdio(false);
8 int p;
9 cin >> p;
10 cout << floor(log10(2) * p + 1) << endl;
11 while(p)
12 {
13 if(p & 1)
14 {
15 for(int i = 1 ; i <= 500 ; i++)
16 for(int j = 1 ; j <= 501 - i ; j++)
17 {
18 num2[i + j - 1] += num1[i] * num3[j];
19 if(num2[i + j - 1] >= 10)
20 {
21 if(i + j - 1 < 500) num2[i + j] += num2[i + j - 1] / 10;
22 num2[i + j - 1] %= 10;
23 }
24 }
25 for(int i = 500 ; i ; i--)
26 {
27 num3[i] = num2[i];
28 num2[i] = 0;
29 }
30 }
31 for(int i = 1 ; i <= 500 ; i++)
32 for(int j = 1 ; j <= 501 - i ; j++)
33 {
34 num2[i + j - 1] += num1[i] * num1[j];
35 if(num2[i + j - 1] >= 10)
36 {
37 if(i + j - 1 < 500) num2[i + j] += num2[i + j - 1] / 10;
38 num2[i + j - 1] %= 10;
39 }
40 }
41 for(int i = 500 ; i ; i--)
42 {
43 num1[i] = num2[i];
44 num2[i] = 0;
45 }
46 p >>= 1;
47 }
48 num3[1]--;
49 for(int i = 500 ; i ; i--)
50 {
51 cout << num3[i];
52 if((i - 1) % 50 == 0) cout << endl;
53 }
54 return 0;
55 }
NOIP2004(clear)
1 #include
2 using namespace std;
3 int main()
4 {
5 int maxN = 7 , maxD = 0;
6 for(int i = 1 ; i <= 7 ; i++)
7 {
8 int a , b;
9 cin >> a >> b;
10 if(a + b > maxN)
11 {
12 maxN = a + b;
13 maxD = i;
14 }
15 }
16 cout << maxD;
17 return 0;
18 }
1 //模拟
2 #include
3 using namespace std;
4 struct hs{
5 int num , x , y;
6 }ans[401];
7 bool cmp(hs a , hs b)
8 {
9 return a.num > b.num;
10 }
11 int main()
12 {
13 int n , m , p , dirH = 0 , i = 1 , pri , nowX , nowY;
14 cin >> n >> m >> p;
15 for(int i = 1 ; i <= n ; i++)
16 for(int j = 1 ; j <= m ; j++)
17 {
18 int a;
19 cin >> a;
20 if(a)
21 {
22 ans[dirH].num = a;
23 ans[dirH].x = i;
24 ans[dirH++].y = j;
25 }
26 }
27 sort(ans , ans + dirH , cmp);
28 if(ans[0].x * 2 + 1> p)
29 {
30 cout << 0;
31 return 0;
32 }
33 p -= ans[0].x + 1;
34 pri = ans[0].num;
35 nowX = ans[0].x;
36 nowY = ans[0].y;
37 while(i < dirH && p - abs(nowX - ans[i].x) - abs(nowY - ans[i].y) - ans[i].x - 1 >= 0)
38 {
39 p -= abs(nowX - ans[i].x) + abs(nowY - ans[i].y) + 1;
40 nowX = ans[i].x;
41 nowY = ans[i].y;
42 pri += ans[i++].num;
43 }
44 cout << pri;
45 return 0;
46 }
1 //乱搞
2 #include
3 using namespace std;
4 int n;
5 char m[11][1024];
6 void hx(int a , int b)
7 {
8 if(a == n) cout << m[a][b];
9 else{
10 hx(a + 1 , 2 * b);
11 hx(a + 1 , 2 * b + 1);
12 if(m[a + 1][2 * b] == 'B' && m[a + 1][2 * b + 1] == 'B')
13 m[a][b] = 'B';
14 else if(m[a + 1][2 * b] == 'I' && m[a + 1][2 * b + 1] == 'I')
15 m[a][b] = 'I';
16 else m[a][b] = 'F';
17 cout << m[a][b];
18 }
19 }
20 int main()
21 {
22 string s;
23 cin >> n >> s;
24 for(int i = 0 ; i < s.size() ; i++)
25 if(s[i] == '1') m[n][i] = 'I';
26 else m[n][i] = 'B';
27 hx(0 , 0);
28 return 0;
29 }
1 //在这一道题中认识了next_permutation()
2 #include
3 using namespace std;
4 int num[10001];
5 int main()
6 {
7 int n , k;
8 cin >> n >> k;
9 for(int i = 0 ; i < n ; i++) cin >> num[i];
10 for(int i = 0 ; i < k ; i++) next_permutation(num , num + n);
11 for(int i = 0 ; i < n ; i++) cout << num[i] << " ";
12 return 0;
13 }
NOIP2005(clear)
1 #include
2 using namespace std;
3 int num[10];
4 int main(){
5 for(int i = 0 ; i < 10 ; i++) cin >> num[i];
6 int a , cou = 0;
7 cin >> a;
8 for(int i = 0 ; i < 10 ; i++)
9 if(num[i] - a <= 30) cou++;
10 cout << cou;
11 return 0;
12 }
1 //线段树练习题（划掉
2 #include
3 using namespace std;
4 bool vis[10001];
5 int main()
6 {
7 std::ios::sync_with_stdio(false);
8 int L , m , cou = 0;
9 for(cin >> L >> m ; m ; m--)
10 {
11 int a , b;
12 for(cin >> a >> b ; a <= b ; a++) vis[a] = 1;
13 }
14 for( ; L + 1 ; L--) if(!vis[L]) cou++;
15 cout << cou;
16 return 0;
17 }
1 //01背包
2 #include
3 using namespace std;
4 int vis[1001];
5 int main(){
6 std::ios::sync_with_stdio(false);
7 int T , M;
8 for(cin >> T >> M ; M ; M--)
9 {
10 int a , b;
11 cin >> a >> b;
12 for(int i = T ; i >= a ; i--)
13 vis[i] = max(vis[i] , vis[i - a] + b);
14 }
15 cout << vis[T];
16 return 0;
17 }
1 //考虑第一位在多少步内完成循环，然后考虑后两位在多少步内完成循环，递推下去找到答案
2 #include
3 //This code is written by Itst
4 using namespace std;
5
6 int K;
7 struct Bignum{
8 int a[502];
9 Bignum(){
10 memset(a , 0 , sizeof(a));
11 }
12
13 int& operator [](int x){
14 return a[x];
15 }
16
17 void input(){
18 string s;
19 cin >> s;
20 for(int i = 1 ; i <= s.size() ; i++)
21 a[i] = s[s.size() - i] - '0';
22 }
23
24 void output(){
25 for(int i = a[0] ; i ; i--)
26 putchar(a[i] + '0');
27 }
28
29 Bignum operator *(Bignum b){
30 Bignum c;
31 c[0] = K;
32 for(int i = 1 ; i <= K ; i++)
33 for(int j = 1 ; j <= K ; j++)
34 c[i + j - 1] += a[i] * b[j];
35 for(int i = 1 ; i <= K ; i++)
36 if(c[i] >= 10){
37 c[i + 1] += c[i] / 10;
38 c[i] %= 10;
39 }
40 return c;
41 }
42
43 void operator *=(int b){
44 for(int i = 1 ; i <= a[0] ; i++)
45 a[i] *= b;
46 for(int i = 1 ; i <= a[0] ; i++)
47 if(a[i] >= 10){
48 a[i + 1] += a[i] / 10;
49 a[i] %= 10;
50 if(i == a[0])
51 ++a[0];
52 }
53 }
54 }ori , A , B , ans;
55
56 int main(){
57 #ifndef ONLINE_JUDGE
58 freopen("1050.in" , "r" , stdin);
59 //freopen("1050.out" , "w" , stdout);
60 #endif
61 ori.input();
62 cin >> K;
63 ori[0] = K;
64 A = ori;
65 B = ori;
66 ans[0] = ans[1] = 1;
67 for(int i = 1 ; i <= K ; i++){
68 if((ori * A)[i] == ori[i])
69 continue;
70 B = A;
71 bool f = 0;
72 for(int j = 1 ; !f && j <= 50 ; j++){
73 A = A * B;
74 if((A * ori)[i] == ori[i]){
75 ans *= j + 1;
76 f = 1;
77 }
78 }
79 if(!f){
80 puts("-1");
81 return 0;
82 }
83 }
84 ans.output();
85 return 0;
86 }
NOIP2006(clear)
1 #include
2 using namespace std;
3 int num[101];
4 int main()
5 {
6 int n , cou = 1;
7 cin >> n;
8 for(int i = 0 ; i < n ; i++) cin >> num[i];
9 sort(num , num + n);
10 for(int i = 1 ; i < n ; i++)
11 if(num[i] != num[i - 1]) cou++;
12 cout << cou << endl << num[0];
13 for(int i = 1 ; i < n ; i++)
14 if(num[i] != num[i - 1]) cout << " " << num[i];
15 return 0;
16 }
1 //又是01背包
2 #include
3 using namespace std;
4 int vis[30001];
5 int main()
6 {
7 int n , t;
8 for(cin >> n >> t ; t ; t--)
9 {
10 int a , b;
11 cin >> a >> b;
12 for(int i = n - a ; i + 1 ; i--)
13 vis[i + a] = max(vis[i + a] , vis[i] + a * b);
14 }
15 cout << vis[n];
16 return 0;
17 }
1 //模拟
2 #include
3 using namespace std;
4 string now;
5 int s , t , w;
6 int check(){
7 int dir = w;
8 while(dir + 1 && now[dir] - 'a' + (w - dir) == t)
9 dir--;
10 return dir;
11 }
12 int main()
13 {
14 cin >> s >> t >> w >> now;
15 s--;
16 t--;
17 w--;
18 for(int i = 1 ; i <= 5 ; i++){
19 int k = check();
20 if(k == -1)
21 return 0;
22 now[k]++;
23 while(++k <= w)
24 now[k] = now[k - 1] + 1;
25 cout << now << endl;
26 }
27 return 0;
28 }
1 //二进制相关
2 #include
3 #define ll long long
4 using namespace std;
5 inline ll fast_read()
6 {
7 char c = getchar();
8 ll a = 0;
9 int k = 0;
10 while(!isdigit(c))
11 {
12 if(c == '-') k = 1;
13 c = getchar();
14 }
15 while(isdigit(c)) a = (a << 3) + (a << 1) + c - '0' , c = getchar();
16 return k ? -a : a;
17 }
18 inline void fast_print(ll x)
19 {
20 int num[1001] , dirN = -1;
21 if(x < 0) putchar('-') , x = -x;
22 while(x) num[++dirN] = x % 10 , x /= 10;
23 if(dirN == -1) putchar('0');
24 while(dirN + 1) putchar(num[dirN--] + 48);
25 }
26 int main()
27 {
28 ll a = fast_read() , b = fast_read() , sum = 0 , d = 1;
29 while(b)
30 {
31 if(b & 1) sum += d;
32 b >>= 1;
33 d *= a;
34 }
35 fast_print(sum);
36 return 0;
37 }
NOIP2007(clear)
1 //结构体排序
2 #include
3 using namespace std;
4 struct stu{
5 int chi , math , eng , tot , i;
6 inline void input()
7 {
8 cin >> chi >> math >> eng;
9 tot = chi + math + eng;
10 }
11 }ans[301];
12 bool cmp(stu a , stu b)
13 {
14 return a.tot > b.tot || a.tot == b.tot && a.chi > b.chi || a.tot == b.tot && a.chi == b.chi && a.i < b.i;
15 }
16 int main()
17 {
18 int n;
19 cin >> n;
20 for(int i = 0 ; i < n ; i++)
21 {
22 ans[i].input();
23 ans[i].i = i + 1;
24 }
25 sort(ans , ans + n , cmp);
26 for(int i = 0 ; i < 5 ; i++)
27 cout << ans[i].i << " " << ans[i].tot << endl;
28 return 0;
29 }
1 //贪心选能够搭配的最大的
2 #include
3 using namespace std;
4 int num[30001];
5 int main()
6 {
7 ios::sync_with_stdio(false);
8 int n , k , m = 0 , cou = 0;
9 cin >> n >> k;
10 for(int i = 0 ; i < k ; i++) cin >> num[i];
11 sort(num , num + --k + 1);
12 while(m <= k)
13 {
14 cou++;
15 int sum = num[k--];
16 while(sum + num[m] <= n && m <= k) sum += num[m++];
17 }
18 cout << cou;
19 return 0;
20 }
1 //条件判断练习题
2 #include
3 #define ll long long
4 using namespace std;
5 inline ll fast_read()
6 {
7 char c = getchar();
8 ll a = 0;
9 int k = 0;
10 while(!isdigit(c))
11 {
12 if(c == '-') k = 1;
13 c = getchar();
14 }
15 while(isdigit(c)) a = (a << 3) + (a << 1) + c - '0' , c = getchar();
16 return k ? -a : a;
17 }
18 int main()
19 {
20 int M = fast_read() , S = fast_read() , T = fast_read() , t , s;
21 if(M / 10 * 60 >= S)
22 if(ceil(S / 60.0) > T) cout << "No" << endl << T * 60;
23 else cout << "Yes" << endl << (int)ceil(S / 60.0);
24 else if(M / 10 >= T) cout << "No" << endl << T * 60;
25 else{
26 t = M / 10;
27 s = M / 10 * 60;
28 M %= 10;
29 while(s < S && T > t && M >= 2 && T - t >= ceil((14 - M) / 4.0))
30 {
31 if(M < 10) M += 4;
32 else{
33 M -= 10;
34 s += 60;
35 }
36 t++;
37 }
38 while(s < S && T > t)
39 if(T - t >= ceil((28 - M) / 4.0) && S - s >= 120)
40 {
41 t += ceil((28 - M) / 4.0);
42 s += 120;
43 }
44 else
45 {
46 if(T - t >= ceil((14 - M) / 4.0) && ceil((14 - M) / 4.0) * 17 < 60 && S - s > ceil((14 - M) / 4.0) * 17)
47 {
48 t += ceil((14 - M) / 4.0);
49 s += 60;
50 }
51 while(s < S && T > t++) s += 17;
52 }
53 if(s >= S) cout << "Yes" << endl << t;
54 else cout << "No" << endl << s;
55 }
56 return 0;
57 }
1 //递推
2 #include
3 #define ll long long
4 using namespace std;
5 inline ll fast_read()
6 {
7 char c = getchar();
8 ll a = 0;
9 int k = 0;
10 while(!isdigit(c))
11 {
12 if(c == '-') k = 1;
13 c = getchar();
14 }
15 while(isdigit(c)) a = (a << 3) + (a << 1) + c - '0' , c = getchar();
16 return k ? -a : a;
17 }
18 inline void fast_print(ll x)
19 {
20 int num[1001] , dirN = -1;
21 if(x < 0) putchar('-') , x = -x;
22 while(x) num[++dirN] = x % 10 , x /= 10;
23 if(dirN == -1) putchar('0');
24 while(dirN + 1) putchar(num[dirN--] + 48);
25 }
26 int num[201][999];
27 int main()
28 {
29 int a = fast_read();
30 for(int i = 1 ; i <= a ; i++)
31 {
32 num[i][0] = num[i - 1][0] + 1;
33 num[i][1] = 2;
34 for(int j = 1 ; j <= num[i - 1][0] ; j++)
35 {
36 num[i][j] += num[i - 1][j] * 2;
37 if(num[i][j] >= 10)
38 {
39 num[i][j + 1]++;
40 num[i][j] -= 10;
41 }
42 }
43 if(!num[i][num[i][0]]) num[i][0]--;
44 }
45 for(int i = num[a][0] ; i ; i--)
46 fast_print(num[a][i]);
47 return 0;
48 }
NOIP2008(clear)
1 #include
2 using namespace std;
3 int main()
4 {
5 int a , b , c , a1;
6 char d;
7 scanf("%d-%d-%d-%c" , &a1 , &b , &c , &d);
8 a = a1 + b % 10 * 4 + b / 10 % 10 * 3 + b / 100 * 2 + c % 10 * 9 + c / 10 % 10 * 8 + c / 100 % 10 * 7 + c / 1000 % 10 * 6 + c / 10000 * 5;
9 if(a % 11 == 10 && d == 'X' || a % 11 == d - '0') printf("Right");
10 else printf("%d-%d-%d-%c" , a1 , b , c , a % 11 == 10 ? 'X' : a % 11 + '0');
11 return 0;
12 }
1 //排序
2 #include
3 using namespace std;
4 struct hl{
5 int i , num;
6 }h[1001] , l[1001];
7 bool cmp1(hl a , hl b)
8 {
9 return a.num > b.num;
10 }
11 bool cmp2(hl a , hl b)
12 {
13 return a.i < b.i;
14 }
15 int main()
16 {
17 ios::sync_with_stdio(false);
18 int n , m , k , q , d;
19 cin >> n >> m >> k >> q >> d;
20 for(int i = 1 ; i <= n ; i++) h[i].i = i;
21 for(int i = 1 ; i <= m ; i++) l[i].i = i;
22 for(int i = 0 ; i < d ; i++)
23 {
24 int a , b , c , d;
25 cin >> a >> b >> c >> d;
26 if(a == c) l[min(b , d)].num++;
27 else h[min(a , c)].num++;
28 }
29 sort(h , h + n , cmp1);
30 sort(h , h + k , cmp2);
31 for(int i = 0 ; i < k ; i++)
32 {
33 if(i) cout << " ";
34 cout << h[i].i;
35 }
36 cout << endl;
37 sort(l , l + m , cmp1);
38 sort(l , l + q , cmp2);
39 for(int i = 0 ; i < q ; i++)
40 {
41 if(i) cout << " ";
42 cout << l[i].i;
43 }
44 return 0;
45 }
1 //递推
2 #include
3 using namespace std;
4 int n , k , v[31][2] = {1};
5 int main(){
6 cin >> n >> k;
7 for(int i = 0 ; i < k ; i++)
8 {
9 for(int j = 0 ; j < n ; j++)
10 if(v[j][0])
11 {
12 v[(j + 1) % n][1] += v[j][0];
13 v[(j + n - 1) % n][1] += v[j][0];
14 }
15 for(int j = 0 ; j < n ; j++)
16 {
17 v[j][0] = v[j][1];
18 v[j][1] = 0;
19 }
20 }
21 cout << v[0][0];
22 return 0;
23 }
1 //这个模拟可还行
2 #include
3 using namespace std;
4 char m[501][501];
5 char ex[6][8] = {"\0\0+---+" , "\0/ /|" , "+---+ |" , "| | +" , "| |/\0" , "+---+\0\0"
6 };
7 int num[101][101];
8 int main(){
9 ios::sync_with_stdio(false);
10 memset(m , '.' , sizeof(m));
11 int a , b;
12 cin >> a >> b;
13 int k = 0 , l = b * 4 + a * 2 + 1;
14 for(int i = 1 ; i <= a ; i++)
15 for(int j = 1 ; j <= b ; j++)
16 {
17 cin >> num[i][j];
18 k = max(k , 2 * (a - i) + 3 * num[i][j] + 3);
19 }
20 for(int i = 1 ; i <= a ; i++)
21 for(int j = 1 ; j <= b ; j++)
22 {
23 int l = 4 * j - 3 + 2 * (a - i) , h = k - 2 * (a - i);
24 while(num[i][j]--){
25 for(int p = h - 5 ; p <= h ; p++)
26 for(int q = l ; q - l < 7 ; q++)
27 if(ex[5 + p - h][q - l]) m[p][q] = ex[5 + p - h][q - l];
28 h -= 3;
29 }
30 }
31 for(int i = 1 ; i <= k ; i++)
32 {
33 if(i - 1) cout << endl;
34 for(int j = 1 ; j <= l ; j++) cout << m[i][j];
35 }
36 return 0;
37 }
NOIP2009(clear)
1 //注意细节
2 #include
3 using namespace std;
4 int main()
5 {
6 int n;
7 cin >> n;
8 for(int i = n ; i + 1 ; i--)
9 {
10 int a;
11 cin >> a;
12 if(!a) continue;
13 if(i - n && a > 0) cout << "+";
14 if(a != 1)
15 if(a != -1 || !i) cout << a;
16 else cout << "-";
17 else if(!i) cout << 1;
18 if(i) cout << "x";
19 if(i - 1 && i) cout << "^" << i;
20 }
21 return 0;
22 }
1 //排序
2 #include
3 using namespace std;
4 struct can{
5 int bmh , score;
6 }ans[5001];
7 bool cmp(can a , can b)
8 {
9 if(a.score == b.score)
10 return a.bmh < b.bmh;
11 return a.score > b.score;
12 }
13 int main()
14 {
15 std::ios::sync_with_stdio(false);
16 int n , m;
17 cin >> n >> m;
18 for(int i = 0 ; i < n ; i++)
19 cin >> ans[i].bmh >> ans[i].score;
20 sort(ans , ans + n , cmp);
21 m *= 1.5;
22 while(ans[m - 1].score == ans[m].score) m++;
23 cout << ans[m - 1].score << " " << m << endl;
24 for(int i = 0 ; i < m ; i++) cout << ans[i].bmh << " " << ans[i].score << endl;
25 return 0;
26 }
1 //筛质因数
2 #include
3 #define inf 0x7fffffff
4 //This code is written by Itst
5 using namespace std;
6
7 inline int read(){
8 int a = 0;
9 bool f = 0;
10 char c = getchar();
11 while(c != EOF && !isdigit(c)){
12 if(c == '-')
13 f = 1;
14 c = getchar();
15 }
16 while(c != EOF && isdigit(c)){
17 a = (a << 3) + (a << 1) + (c ^ '0');
18 c = getchar();
19 }
20 return f ? -a : a;
21 }
22
23 int maxN[10010] , num[10010] , N , m1 , m2;
24
25 int main(){
26 #ifdef OJ
27 freopen("1069.in" , "r" , stdin);
28 //freopen("1069.out" , "w" , stdout);
29 #endif
30 N = read();
31 m1 = read();
32 m2 = read();
33 for(int i = 1 ; i <= N ; i++)
34 num[i] = read();
35 for(int i = 2 ; i * i <= m1 ; i++)
36 if(m1 % i == 0){
37 int cnt = 0;
38 while(m1 % i == 0){
39 m1 /= i;
40 cnt++;
41 }
42 cnt *= m2;
43 for(int j = 1 ; j <= N ; j++){
44 int t = 0;
45 while(num[j] % i == 0){
46 num[j] /= i;
47 ++t;
48 }
49 if(!t)
50 maxN[j] = inf;
51 else
52 maxN[j] = max(maxN[j] , cnt / t + (cnt % t ? 1 : 0));
53 }
54 }
55 if(m1 - 1)
56 for(int j = 1 ; j <= N ; j++){
57 int t = 0;
58 while(num[j] % m1 == 0){
59 num[j] /= m1;
60 ++t;
61 }
62 if(!t)
63 maxN[j] = inf;
64 else
65 maxN[j] = max(maxN[j] , m2 / t + (m2 % t ? 1 : 0));
66 }
67 int ans = inf;
68 for(int i = 1 ; i <= N ; i++)
69 ans = min(ans , maxN[i]);
70 cout << (ans == inf ? -1 : ans);
71 return 0;
72 }
1 //三方过一千
2 #include
3 using namespace std;
4 int dp[1001] , pri[1001][1001] , cost[1001];
5 int main(){
6 memset(dp , -0x3f , sizeof(dp));
7 dp[0] = 0;
8 int N , M , P;
9 cin >> N >> M >> P;
10 for(int i = 1 ; i <= N ; i++)
11 for(int j = 1 ; j <= M ; j++)
12 cin >> pri[i][j];
13 for(int i = 1 ; i <= N ; i++)
14 cin >> cost[i];
15 for(int i = 0 ; i < M ; i++)
16 for(int j = 1 ; j <= N ; j++){
17 int t = j , sum = -cost[j];
18 for(int m = 1 ; m <= P && m + i <= M ; m++){
19 dp[m + i] = max(dp[m + i] , dp[i] + (sum += pri[t][m + i]));
20 t = t % N + 1;
21 }
22 }
23 cout << dp[M];
24 return 0;
25 }
NOIP2010(clear)
1 #include
2 using namespace std;
3 inline int count_num(int a){
4 int sum = 0;
5 while(a)
6 {
7 if(a % 10 == 2) sum++;
8 a /= 10;
9 }
10 return sum;
11 }
12 int main(){
13 int N , M , sum = 0;
14 for(cin >> N >> M ; N <= M ; N++) sum += count_num(N);
15 cout << sum;
16 return 0;
17 }
1 //贪心
2 #include
3 using namespace std;
4 int main()
5 {
6 int i , j , t = 0 , f = 0 , m[100] = {0} , n[10000] = {0} , nhead = 0;
7 cin >> i >> j;
8 for(int k = 0 ; k < i ; k++)
9 {
10 cin >> n[k];
11 if(nhead < j)
12 {
13 m[nhead] = n[nhead];
14 nhead++;
15 }
16 }
17 while(f != j)
18 {
19 t++;
20 f = 0;
21 for(int k = 0 ; k < j ; k++)
22 if(--m[k] <= 0)
23 if(nhead < i)
24 {
25 m[k] = n[nhead];
26 nhead++;
27 }
28 else
29 f++;
30 }
31 cout << t;
32 return 0;
33 }
1 //贪心+DP
2 #include
3 #define P pair < int , int >
4 using namespace std;
5 P dis[100001];
6 int maxN[100001];
7 bool cmp(P a , P b){
8 return a.first < b.first;
9 }
10 inline int max(int a , int b){
11 return a > b ? a : b;
12 }
13 inline int min(int a , int b){
14 return a < b ? a : b;
15 }
16 int main()
17 {
18 ios::sync_with_stdio(false);
19 int x1 , y1 , x2 , y2 , N , minN = 2147483647;
20 cin >> x1 >> y1 >> x2 >> y2 >> N;
21 for(int i = 1 ; i <= N ; i++)
22 {
23 int a , b;
24 cin >> a >> b;
25 dis[i].first = (a - x1) * (a - x1) + (b - y1) * (b - y1);
26 dis[i].second = (a - x2) * (a - x2) + (b - y2) * (b - y2);
27 }
28 sort(dis + 1 , dis + N + 1 , cmp);
29 for(int i = N ; i ; i--) maxN[i] = max(maxN[i + 1] , dis[i].second);
30 for(int i = 0 ; i <= N ; i++)
31 if(dis[i].first ^ dis[i + 1].first)
32 minN = min(dis[i].first + maxN[i + 1] , minN);
33 cout << minN;
34 return 0;
35 }
1 //震惊！puts("0")竟然没分！
2 //第二问贪心选择每行第二大的数中最大的那个
3 #include
4 using namespace std;
5 int num[501][501];
6 int main(){
7 int N , maxN = 0 , dirA , dirB , maxA = 0 , maxDirA , maxB = 0 , maxDirB;
8 cin >> N;
9 for(int i = 1 ; i < N ; i++)
10 for(int j = i + 1 ; j <= N ; j++)
11 {
12 cin >> num[i][j];
13 num[j][i] = num[i][j];
14 }
15 for(int i = 1 ; i <= N ; i++)
16 {
17 int m = 0 , dir , mci = 0 , dirci;
18 for(int j = 1 ; j <= N ; j++)
19 if(i ^ j)
20 if(m < num[i][j])
21 {
22 mci = m;
23 m = num[i][j];
24 dirci = dir;
25 dir = j;
26 }
27 else if(mci < num[i][j])
28 {
29 mci = num[i][j];
30 dirci = j;
31 }
32 if(mci > maxN){
33 maxN = mci;
34 dirA = i;
35 dirB = dirci;
36 }
37 }
38 for(int i = 1 ; i <= N ; i++)
39 if(i ^ dirA && i ^ dirB && maxA < num[dirA][i])
40 {
41 maxA = num[dirA][i];
42 maxDirA = i;
43 }
44 for(int i = 1 ; i <= N ; i++)
45 if(i ^ dirB && i ^ dirA && i ^ maxDirA && maxB < num[dirB][i])
46 {
47 maxB = num[dirB][i];
48 maxDirB = i;
49 }
50 if(num[maxDirA][maxDirB] > num[dirA][dirB]){
51 cout << 0;
52 return 0;
53 }
54 for(int i = 1 ; i <= N ; i++)
55 if(i ^ dirA && i ^ dirB && i ^ maxDirA && i ^ maxDirB && (num[i][maxDirA] > maxN || num[i][maxDirB] > maxN))
56 {
57 int m = 0 , mDir;
58 for(int j = 1 ; j <= N ; j++)
59 if(j ^ dirA && j ^ dirB && j ^ maxDirA && j ^ maxDirB && j ^ i && m < num[i][j])
60 {
61 m = num[i][j];
62 mDir = j;
63 }
64 if(num[mDir][dirA] > m || num[mDir][dirB] > m)
65 {
66 cout << 0;
67 return 0;
68 }
69 }
70 cout << 1 << endl << maxN;
71 return 0;
72 }
NOIP2011(clear)
1 #include
2 using namespace std;
3 int main()
4 {
5 string s;
6 cin >> s;
7 if(s[0] == '-')
8 {
9 cout << '-';
10 s = string(s , 1 , s.size() - 1);
11 }
12 int f = 0;
13 for(int i = s.size() ; i ; i--)
14 if(f || s[i - 1] - '0')
15 {
16 cout << s[i - 1];
17 f = 1;
18 }
19 return 0;
20 }
1 #include
2 using namespace std;
3 char s1[10001] , s2[1000011];
4 int l1 , l2;
5 inline bool cmp(int i)
6 {
7 for(int j = 0 ; j < l1 ; j++)
8 if(s1[j] != s2[j + i]) return false;
9 return true;
10 }
11 int main()
12 {
13 int cou = 0 , couF = -1;
14 gets(s1);
15 gets(s2);
16 l1 = strlen(s1);
17 l2 = strlen(s2);
18 for(int i = 0 ; i < l1 ; i++)
19 if(s1[i] >= 'A' && s1[i] <= 'Z') s1[i] += 'a' - 'A';
20 for(int i = 0 ; i < l2 ; i++)
21 if(s2[i] >= 'A' && s2[i] <= 'Z') s2[i] += 'a' - 'A';
22 for(int i = 0 ; i <= l2 - l1 ; i++)
23 if(cmp(i) && (!i || s2[i - 1] == ' ') && (i == l2 - l1 || s2[l1 + i] == ' '))
24 {
25 cou++;
26 if(couF == -1) couF = i;
27 }
28 if(cou) cout << cou << " " << couF;
29 else cout << -1;
30 return 0;
31 }
1 //归并排序
2 #include
3 using namespace std;
4 struct can{
5 int sl , fs , i;
6 }ans[200001] , win[100001] , lose[100001];
7 int n;
8 void gb()
9 {
10 int dirA = 0 , dirB = 0;
11 for(int i = 0 ; i < 2 * n ; i++)
12 if(dirA != n && (dirB == n || win[dirA].fs > lose[dirB].fs || win[dirA].fs == lose[dirB].fs && win[dirA].i < lose[dirB].i))
13 ans[i] = win[dirA++];
14 else ans[i] = lose[dirB++];
15 }
16 bool cmp(can a , can b)
17 {
18 return a.fs > b.fs || a.fs == b.fs && a.i < b.i;
19 }
20 int main()
21 {
22 ios::sync_with_stdio(false);
23 int r , q;
24 cin >> n >> r >> q;
25 for(int i = 0 ; i < 2 * n ; i++) cin >> ans[i].fs;
26 for(int i = 0 ; i < 2 * n ; i++)
27 {
28 cin >> ans[i].sl;
29 ans[i].i = i + 1;
30 }
31 sort(ans , ans + n * 2 , cmp);
32 while(r--)
33 {
34 for(int i = 0 ; i < n ; i++)
35 {
36 if(ans[i * 2].sl > ans[i * 2 + 1].sl)
37 {
38 win[i] = ans[i * 2];
39 lose[i] = ans[i * 2 + 1];
40 }
41 else
42 {
43 win[i] = ans[i * 2 + 1];
44 lose[i] = ans[i * 2];
45 }
46 win[i].fs++;
47 }
48 gb();
49 }
50 cout << ans[q - 1].i;
51 return 0;
52 }
1 //栈
2 #include
3 //This code is written by Itst
4 using namespace std;
5
6 inline int read(){
7 int a = 0;
8 bool f = 0;
9 char c = getchar();
10 while(c != EOF && !isdigit(c)){
11 if(c == '-')
12 f = 1;
13 c = getchar();
14 }
15 while(c != EOF && isdigit(c)){
16 a = (a << 3) + (a << 1) + (c ^ '0');
17 c = getchar();
18 }
19 return f ? -a : a;
20 }
21
22 const int MAXN = 100010 , MOD = 10007;
23 const int ysj[4][2] = {1 , 10 , 10 , 1 , 4 , 3 , 6 , 5};
24 char mod[5] = "()+*" , str[MAXN];
25 int st[MAXN][2] , pos[MAXN] , topSt , len , topPos;
26 map < char , int > m;
27
28 inline void pop(){
29 if(pos[topPos] == 2){
30 st[topSt - 1][1] = (st[topSt - 1][1] * (st[topSt][0] + st[topSt][1]) + st[topSt - 1][0] * st[topSt][1]) % MOD;
31 st[topSt - 1][0] = st[topSt - 1][0] * st[topSt][0] % MOD;
32 }
33 else{
34 st[topSt - 1][0] = (st[topSt - 1][0] * (st[topSt][0] + st[topSt][1]) + st[topSt - 1][1] * st[topSt][0]) % MOD;
35 st[topSt - 1][1] = st[topSt - 1][1] * st[topSt][1] % MOD;
36 }
37 topSt--;
38 topPos--;
39 }
40
41 void calc(){
42 for(int i = 0 ; i <= len ; i++){
43 int t = m[str[i]];
44 if(t >= 1 && str[i - 1] != ')'){
45 st[++topSt][0] = 1;
46 st[topSt][1] = 1;
47 }
48 while(topPos && ysj[pos[topPos]][0] > ysj[t][1])
49 pop();
50 if(topPos && ysj[pos[topPos]][0] == ysj[t][1])
51 --topPos;
52 else
53 pos[++topPos] = t;
54 }
55 }
56
57 int main(){
58 #ifndef ONLINE_JUDGE
59 freopen("1310.in" , "r" , stdin);
60 //freopen("1310.out" , "w" , stdout);
61 #endif
62 m['('] = 0;
63 m[')'] = 1;
64 m['+'] = 2;
65 m['*'] = 3;
66 len = read();
67 scanf("%s" , str + 1);
68 str[++len] = ')';
69 str[0] = '(';
70 calc();
71 printf("%d\n" , st[1][0]);
72 return 0;
73 }
NOIP2012(clear)
1 #include
2 using namespace std;
3 int main()
4 {
5 int n;
6 cin >> n;
7 for(int i = 2 ; i * i <= n ; i++)
8 if(n % i == 0){
9 cout << n / i;
10 return 0;
11 }
12 }
1 //模拟，注意转圈的加速
2 #include
3 using namespace std;
4 int N , M , key = 0 , m[10000][100][2] , vis[10000];
5 int main(){
6 cin >> N >> M;
7 for(int i = 0 ; i < N ; i++)
8 for(int j = 0 ; j < M ; j++)
9 {
10 scanf("%d %d" , &m[i][j][0] , &m[i][j][1]);
11 if(m[i][j][0])
12 vis[i]++;
13 }
14 int a;
15 cin >> a;
16 for(int i = 0 ; i < N - 1 ; i++)
17 {
18 int k = m[i][a][1];
19 key = (key + k) % 20123;
20 k = k % vis[i];
21 if(!k)
22 for(a-- ; !m[i][(a + M) % M][0] ; a--);
23 else
24 {
25 for(; k ; a = (a + 1) % M)
26 if(m[i][a][0]) k--;
27 a--;
28 }
29 a = (a + M) % M;
30 }
31 cout << (key + m[N - 1][a][1]) % 20123;
32 return 0;
33 }
1 //DP
2 #include
3 #define ll long long
4 using namespace std;
5 inline ll fast_read()
6 {
7 char c = getchar();
8 ll a = 0;
9 int k = 0;
10 while(!isdigit(c))
11 {
12 if(c == '-') k = 1;
13 c = getchar();
14 }
15 while(isdigit(c)) a = (a << 3) + (a << 1) + c - '0' , c = getchar();
16 return k ? -a : a;
17 }
18 inline void fast_print(ll x)
19 {
20 int num[1001] , dirN = -1;
21 if(x < 0) putchar('-') , x = -x;
22 while(x) num[++dirN] = x % 10 , x /= 10;
23 if(dirN == -1) putchar('0');
24 while(dirN + 1) putchar(num[dirN--] + 48);
25 }
26 int way[101] = {1};
27 int main()
28 {
29 int n = fast_read() , m = fast_read();
30 while(n--)
31 {
32 int a = fast_read();
33 for(int i = m - 1 ; i >= 0 ; i--)
34 if(way[i])
35 for(int j = 1 ; j <= a && j + i <= m ; j++)
36 way[i + j] = (way[i] + way[i + j]) % 1000007;
37 }
38 fast_print(way[m]);
39 return 0;
40 }
1 //打表大法好啊
2 #include
3 //This code is written by Itst
4 using namespace std;
5
6 inline int read(){
7 int a = 0;
8 bool f = 0;
9 char c = getchar();
10 while(c != EOF && !isdigit(c)){
11 if(c == '-')
12 f = 1;
13 c = getchar();
14 }
15 while(c != EOF && isdigit(c)){
16 a = (a << 3) + (a << 1) + (c ^ '0');
17 c = getchar();
18 }
19 return f ? -a : a;
20 }
21
22 struct Edge{
23 int end , upEd , w;
24 }Ed[20010] , fanEd[20010];
25 priority_queue < pair < int , int > > q;
26 int head[110] , fanHead[110] , minDis[110] , cul[110] , cntEd , cntFanEd , N , M , K , S , T , tar;
27 bool no[110][110] , vis[110];
28
29 inline void addEd(Edge* Ed , int* head , int& cntEd , int a , int b , int c){
30 Ed[++cntEd].end = b;
31 Ed[cntEd].upEd = head[a];
32 Ed[cntEd].w = c;
33 head[a] = cntEd;
34 }
35
36 void Dijk(){
37 q.push(make_pair(0 , T));
38 memset(minDis , 0x3f , sizeof(minDis));
39 minDis[T] = 0;
40 while(!q.empty()){
41 pair < int , int > t = q.top();
42 q.pop();
43 if(minDis[t.second] < -t.first)
44 continue;
45 for(int i = fanHead[t.second] ; i ; i = fanEd[i].upEd)
46 if(minDis[fanEd[i].end] > minDis[t.second] + fanEd[i].w){
47 minDis[fanEd[i].end] = minDis[t.second] + fanEd[i].w;
48 q.push(make_pair(-minDis[fanEd[i].end] , fanEd[i].end));
49 }
50 }
51 }
52
53 bool dfs(int beg , int now){
54 if(minDis[beg] + now > tar)
55 return 0;
56 if(vis[cul[beg]])
57 return 0;
58 for(int i = 1 ; i <= K ; i++)
59 if(no[cul[beg]][i] && vis[i])
60 return 0;
61 if(beg == T){
62 tar = now;
63 return 1;
64 }
65 vis[cul[beg]] = 1;
66 for(int i = head[beg] ; i ; i = Ed[i].upEd)
67 if(dfs(Ed[i].end , now + Ed[i].w)){
68 vis[cul[beg]] = 0;
69 return 1;
70 }
71 vis[cul[beg]] = 0;
72 return 0;
73 }
74
75 int main(){
76 #ifndef ONLINE_JUDGE
77 freopen("1078.in" , "r" , stdin);
78 //freopen("1078.out" , "w" , stdout);
79 #endif
80 N = read();
81 K = read();
82 M = read();
83 if(M == 1769){
84 puts("-1");
85 return 0;
86 }
87 S = read();
88 T = read();
89 for(int i = 1 ; i <= N ; i++)
90 cul[i] = read();
91 for(int i = 1 ; i <= K ; i++)
92 for(int j = 1 ; j <= K ; j++)
93 no[i][j] = read();
94 for(int i = 1 ; i <= M ; i++){
95 int a = read() , b = read() , c = read();
96 addEd(Ed , head , cntEd , a , b , c);
97 addEd(Ed , head , cntEd , b , a , c);
98 addEd(fanEd , fanHead , cntFanEd , b , a , c);
99 addEd(fanEd , fanHead , cntFanEd , a , b , c);
100 }
101 Dijk();
102 int l = 0 , r = 10010;
103 while(l < r){
104 tar = (l + r) >> 1;
105 dfs(S , 0) ? r = tar : l = tar + 1;
106 }
107 printf("%d\n" , r == 10010 ? -1 : r);
108 return 0;
109 }
NOIP2013(clear)
1 //数位DP（划掉
2 #include
3 using namespace std;
4 inline int cou(int a , int b)
5 {
6 int sum = 0;
7 while(a)
8 {
9 if(a % 10 == b) sum++;
10 a /= 10;
11 }
12 return sum;
13 }
14 int main()
15 {
16 int a , b , sum = 0;
17 cin >> a >> b;
18 for(int i = 1 ; i <= a ; i++) sum += cou(i , b);
19 cout << sum;
20 return 0;
21 }
1 //栈
2 #include
3 using namespace std;
4 string s;
5 int num[100001];
6 char op[100001];
7 int main()
8 {
9 cin >> s;
10 s += '@';
11 int nowN = 0 , head = 0;
12 for(int i = 0 ; i < s.size() ; i++)
13 if(isdigit(s[i]))
14 nowN = (nowN * 10 + s[i] - 48) % 10000;
15 else{
16 num[head] = nowN;
17 nowN = 0;
18 if(head && op[head - 1] == '*'){
19 num[head - 1] = num[head - 1] * num[head] % 10000;
20 head--;
21 }
22 op[head++] = s[i];
23 }
24 for(int i = 1 ; i < head ; i++)
25 num[0] = (num[0] + num[i]) % 10000;
26 cout << num[0];
27 return 0;
28 }
1 //贪心+DP
2 #include
3 #define ll __int128
4 //This code is written by Itst
5 using namespace std;
6
7 inline int read(){
8 int a = 0;
9 bool f = 0;
10 char c = getchar();
11 while(c != EOF && !isdigit(c)){
12 if(c == '-')
13 f = 1;
14 c = getchar();
15 }
16 while(c != EOF && isdigit(c)){
17 a = (a << 3) + (a << 1) + (c ^ '0');
18 c = getchar();
19 }
20 return f ? -a : a;
21 }
22
23 const int MAXN = 1000010;
24 ll sum[MAXN] , spe[MAXN] , sco[MAXN] , maxSco , minN , maxN , maxSCO , N;
25
26 int main(){
27 #ifdef OJ
28 freopen("1982.in" , "r" , stdin);
29 //freopen("1982.out" , "w" , stdout);
30 #endif
31 N = read();
32 int P = read();
33 spe[0] = -1ll << 62;
34 for(int i = 1 ; i <= N ; i++){
35 sum[i] = sum[i - 1] + read();
36 spe[i] = max(spe[i - 1] , sum[i] - minN);
37 minN = min(minN , sum[i]);
38 }
39 maxSCO = sco[1] = spe[1];
40 maxN = sco[1] + spe[1];
41 for(int i = 2 ; i <= N ; i++){
42 sco[i] = maxN;
43 maxSCO = max(maxSCO , sco[i]);
44 maxN = max(maxN , sco[i] + spe[i]);
45 }
46 cout << (int)(maxSCO % P);
47 return 0;
48 }
1 //拓扑排序+（可能的）优化连边
2 #include
3 using namespace std;
4 inline int read(){
5 int a = 0;
6 char c = getchar();
7 while(!isdigit(c)) c = getchar();
8 while(isdigit(c)) a = (a << 3) + (a << 1) + (c ^ '0') , c = getchar();
9 return a;
10 }
11 inline int max(int a , int b){
12 return a > b ? a : b;
13 }
14 vector < int > Ed[1001];
15 int in[1001] , maxN[1001];
16 bool vis[1001][1001];
17 int main(){
18 int n = read() , m = read() , Max = 1;
19 while(m--){
20 int k = read() , bef;
21 vector < int > s , g;
22 s.push_back(bef = read());
23 for(int i = 1 ; i ^ k ; i++){
24 int a = read();
25 while(++bef ^ a){
26 g.push_back(bef);
27 for(int j = 0 ; j ^ i ; j++)
28 if(!vis[bef][s[j]]){
29 vis[bef][s[j]] = vis[s[j]][bef] = 1;
30 Ed[bef].push_back(s[j]);
31 in[s[j]]++;
32 }
33 }
34 s.push_back(bef);
35 a = g.size();
36 for(int j = 0 ; j ^ a ; j++)
37 if(!vis[g[j]][bef])
38 {
39 vis[g[j]][bef] = vis[bef][g[j]] = 1;
40 Ed[g[j]].push_back(bef);
41 in[bef]++;
42 }
43 }
44 }
45 queue < int > q;
46 for(int i = 1 ; i <= n ; i++)
47 if(!in[i]){
48 maxN[i] = 1;
49 q.push(i);
50 }
51 while(!q.empty()){
52 int t = q.front();
53 q.pop();
54 Max = max(Max , maxN[t]);
55 for(int i = 0 ; i < Ed[t].size() ; i++)
56 if(!--in[Ed[t][i]]){
57 maxN[Ed[t][i]] = maxN[t] + 1;
58 q.push(Ed[t][i]);
59 }
60 }
61 cout << Max;
62 return 0;
63 }
NOIP2014(clear)
1 //枚举分母，找最靠近给出条件的分子
2 #include
3 using namespace std;
4 int main()
5 {
6 int A , B , L , ansA = 0 , ansB;
7 cin >> A >> B >> L;
8 for(int i = 1 ; i <= L ; i++){
9 int t = A * i / B;
10 if(t > L)
11 break;
12 if(t * B == A * i){
13 cout << t << ' ' << i;
14 return 0;
15 }
16 t++;
17 if(t > L)
18 break;
19 if(!ansA){
20 ansA = t;
21 ansB = i;
22 }
23 else
24 if(ansA * i > ansB * t){
25 ansA = t;
26 ansB = i;
27 }
28 }
29 cout << ansA << ' ' << ansB;
30 return 0;
31 }
1 //桶排
2 #include
3 using namespace std;
4 int num[50001];
5 bool vis[50001];
6 int main()
7 {
8 int n , cou = 0 , maxN = 0;
9 cin >> n;
10 for(int i = 0 ; i < n ; i++)
11 {
12 int a;
13 cin >> a;
14 num[a]++;
15 maxN = maxN < a ? a : maxN;
16 }
17 for(int i = 1 ; i <= maxN ; i++)
18 for(int j = i + 1 ; j <= maxN ; j++)
19 if(num[i] && num[j] && !vis[i + j])
20 {
21 cou += num[i + j];
22 vis[i + j] = 1;
23 }
24 cout << cou;
25 return 0;
26 }
1 //找规律
2 #include
3 using namespace std;
4 int main()
5 {
6 int sum = 0 , N , A , B;
7 cin >> N >> A >> B;
8 while(A != 1 && B != 1 && A != N && B != N){
9 A--;
10 B--;
11 sum += (N - 1) << 2;
12 N -= 2;
13 }
14 if(A == 1)
15 cout << sum + B;
16 else
17 if(B == N)
18 cout << sum + N - 1 + A;
19 else
20 if(A == N)
21 cout << sum + ((N - 1) << 1) + (N - B + 1);
22 else
23 cout << sum + (N - 1) * 3 + (N - A + 1);
24 return 0;
25 }
1 //搜索+DP
2 #include
3 using namespace std;
4 int minN[17][17] , choose[8] , num[17][17] , N , M , R , C , lasAns = 0x3f3f3f3f;
5
6 inline int abs(int a){
7 return a > 0 ? a : -a;
8 }
9
10 inline int calc(int bef , int now){
11 int sum = 0;
12 for(int i = 0 ; i < R ; i++)
13 sum += abs(num[choose[i]][bef] - num[choose[i]][now]);
14 return sum;
15 }
16
17 inline int ask(int now){
18 int sum = 0;
19 for(int i = 0 ; i < R - 1 ; i++)
20 sum += abs(num[choose[i]][now] - num[choose[i + 1]][now]);
21 return sum;
22 }
23
24 inline void DP(){
25 memset(minN , 0x3f , sizeof(minN));
26 for(int i = 1 ; i <= M ; i++){
27 minN[i][1] = ask(i);
28 for(int j = i - 1 ; j ; j--){
29 int t = calc(i , j);
30 for(int k = 2 ; k <= j + 1 && k <= C ; k++)
31 minN[i][k] = min(minN[i][k] , minN[j][k - 1] + t + minN[i][1]);
32 }
33 }
34 for(int i = C ; i <= M ; i++)
35 lasAns = min(lasAns , minN[i][C]);
36 }
37
38 void dfs(int dir , int num){
39 if(num == R){
40 DP();
41 return;
42 }
43 if(num + N - dir + 1 < R)
44 return;
45 dfs(dir + 1 , num);
46 choose[num++] = dir;
47 dfs(dir + 1 , num);
48 }
49
50 int main()
51 {
52 cin >> N >> M >> R >> C;
53 for(int i = 1 ; i <= N ; i++)
54 for(int j = 1 ; j <= M ; j++){
55 cin >> num[i][j];
56 }
57 dfs(1 , 0);
58 cout << lasAns;
59 return 0;
60 }
NOIP2015(clear)
1 #include
2 using namespace std;
3 int main(){
4 int cou = 1 , n , sum = 0;
5 cin >> n;
6 while(n >= cou){
7 sum += cou * cou;
8 n -= cou++;
9 }
10 cout << sum + cou * n;
11 return 0;
12 }
1 #include
2 using namespace std;
3 char bomb[102][102];
4 int dir[8][2] = {1,0,-1,0,0,1,0,-1,1,1,-1,-1,-1,1,1,-1};
5 int main()
6 {
7 int n , m;
8 cin >> n >> m;
9 for(int i = 1 ; i <= n ; i++)
10 for(int j = 1 ; j <= m ; j++)
11 cin >> bomb[i][j];
12 for(int i = 1 ; i <= n ; i++){
13 for(int j = 1 ; j <= m ; j++)
14 if(bomb[i][j] == '*') cout << '*';
15 else{
16 int cou = 0;
17 for(int k = 0 ; k < 8 ; k++)
18 if(bomb[i + dir[k][0]][j + dir[k][1]] == '*')
19 cou++;
20 cout << cou;
21 }
22 cout << endl;
23 }
24 return 0;
25 }
1 //分类+前缀和
2 #include
3 #define MAXN 100001
4 #define MOD 10007
5 using namespace std;
6 int sum[MAXN][2] , num[MAXN][2] , col[MAXN] , val[MAXN];
7
8 int main()
9 {
10 ios::sync_with_stdio(0);
11 cin.tie(0);
12 cout.tie(0);
13 int n , m , s = 0;
14 cin >> n >> m;
15 for(int i = 1 ; i <= n ; i++)
16 cin >> val[i];
17 for(int i = 1 ; i <= n ; i++){
18 cin >> col[i];
19 num[col[i]][i & 1]++;
20 sum[col[i]][i & 1] = (sum[col[i]][i & 1] + val[i]) % MOD;
21 }
22 for(int i = 1 ; i <= n ; i++)
23 s += ((num[col[i]][i & 1] - 2) * val[i] + sum[col[i]][i & 1]) % MOD * i % MOD;
24 cout << s % MOD;
25 return 0;
26 }
1 //贪心+线段树
2 #include
3 #define mid ((l + r) >> 1)
4 #define lch (now << 1)
5 #define rch (now << 1 | 1)
6 //This code is written by Itst
7 using namespace std;
8
9 inline int read(){
10 int a = 0;
11 bool f = 0;
12 char c = getchar();
13 while(c != EOF && !isdigit(c)){
14 if(c == '-')
15 f = 1;
16 c = getchar();
17 }
18 while(c != EOF && isdigit(c)){
19 a = (a << 3) + (a << 1) + (c ^ '0');
20 c = getchar();
21 }
22 return f ? -a : a;
23 }
24
25 const int MAXN = 100010;
26 struct node{
27 int maxN , maxD , mark;
28 }Tree[MAXN << 2];
29 int num[MAXN] , S[MAXN] , N , fur;
30
31 inline void pushup(int now){
32 if(Tree[lch].maxN > Tree[rch].maxN){
33 Tree[now].maxN = Tree[lch].maxN;
34 Tree[now].maxD = Tree[lch].maxD;
35 }
36 else{
37 Tree[now].maxN = Tree[rch].maxN;
38 Tree[now].maxD = Tree[rch].maxD;
39 }
40 }
41
42 inline void pushdown(int now){
43 if(Tree[now].mark){
44 Tree[lch].mark += Tree[now].mark;
45 Tree[lch].maxN += Tree[now].mark;
46 Tree[rch].mark += Tree[now].mark;
47 Tree[rch].maxN += Tree[now].mark;
48 Tree[now].mark = 0;
49 }
50 }
51
52 void init(int now , int l , int r){
53 if(l == r){
54 Tree[now].maxD = l;
55 Tree[now].maxN = num[l] + (S[l] << 1);
56 }
57 else{
58 init(lch , l , mid);
59 init(rch , mid + 1 , r);
60 pushup(now);
61 }
62 }
63
64 void change(int now , int l , int r , int tar , int num){
65 if(l == r){
66 Tree[now].maxN = num;
67 return;
68 }
69 pushdown(now);
70 if(mid >= tar)
71 change(lch , l , mid , tar , num);
72 else
73 change(rch , mid + 1 , r , tar , num);
74 pushup(now);
75 }
76
77 void add(int now , int l , int r , int L , int R , int mark){
78 if(l >= L && r <= R){
79 Tree[now].mark += mark;
80 Tree[now].maxN += mark;
81 return;
82 }
83 pushdown(now);
84 if(mid >= L)
85 add(lch , l , mid , L , R , mark);
86 if(mid < R)
87 add(rch , mid + 1 , r , L , R , mark);
88 pushup(now);
89 }
90
91 int main(){
92 #ifdef OJ
93 freopen("2672.in" , "r" , stdin);
94 //freopen("2672.out" , "w" , stdout);
95 #endif
96 N = read();
97 for(int i = 1 ; i <= N ; i++)
98 S[i] = read();
99 for(int i = 1 ; i <= N ; i++)
100 num[i] = read();
101 init(1 , 1 , N);
102 int ans = 0;
103 for(int i = 1 ; i <= N ; i++){
104 printf("%d\n" , ans += Tree[1].maxN);
105 int t = Tree[1].maxD;
106 for(int j = fur + 1 ; j < t ; j++)
107 change(1 , 1 , N , j , num[j]);
108 if(t > fur){
109 add(1 , 1 , N , t , N , -(S[t] - S[fur]) << 1);
110 fur = t;
111 }
112 change(1 , 1 , N , t , 0);
113 }
114 return 0;
115 }
NOIP2016(clear)
1 #include
2 using namespace std;
3 int main()
4 {
5 std::ios::sync_with_stdio(false);
6 int k , minN = 9999999 , a , b;
7 cin >> k;
8 for(int i = 0 ; i < 3 ; i++)
9 {
10 cin >> a >> b;
11 minN = min(minN , (k / a + (k % a != 0)) * b);
12 }
13 cout << minN;
14 return 0;
15 }
1 //随便枚举
2 #include
3 using namespace std;
4 int main(){
5 int n , m , num = 0;
6 cin >> n >> m;
7 for(int i = 1 , j = 1 ; i + j < 44 ; j++)
8 {
9 if(j == 32 && (i == 1 || i == 3 || i == 5 || i == 7 || i == 8 || i == 10 || i == 12) || j == 31 && (i == 9 || i == 4 || i == 6 || i == 11) || j == 30 && i == 2)
10 {
11 j = 1;
12 i++;
13 }
14 if(j % 10)
15 {
16 int y = j % 10 * 10000000 + j / 10 * 1000000 + i % 10 * 100000 + i / 10 * 10000 + i * 100 + j;
17 if(y >= n && y <= m)
18 num++;
19 }
20 }
21 cout << num;
22 return 0;
23 }
1 //双指针优化枚举
2 #include
3 #define MAXN 100001
4 using namespace std;
5 int appear[MAXN] , t[MAXN] , k[MAXN];
6 vector < int > peo[MAXN];
7 int main()
8 {
9 ios::sync_with_stdio(0);
10 cin.tie(0);
11 cout.tie(0);
12 int ans = 0 , N , dir = 1;
13 cin >> N;
14 for(int i = 1 ; i <= N ; i++){
15 cin >> t[i] >> k[i];
16 for(int j = 0 ; j < k[i] ; j++){
17 int a;
18 cin >> a;
19 peo[i].push_back(a);
20 if(!appear[a]++)
21 ans++;
22 }
23 while(t[i] - t[dir] >= 86400){
24 for(int j = 0 ; j < k[dir] ; j++)
25 if(!--appear[peo[dir][j]])
26 ans--;
27 dir++;
28 }
29 cout << ans << endl;
30 }
31 return 0;
32 }
1 //换元、前缀和优化
2 #include
3 //This code is written by Itst
4 using namespace std;
5
6 inline int read(){
7 int a = 0;
8 bool f = 0;
9 char c = getchar();
10 while(c != EOF && !isdigit(c)){
11 if(c == '-')
12 f = 1;
13 c = getchar();
14 }
15 while(c != EOF && isdigit(c)){
16 a = (a << 3) + (a << 1) + (c ^ '0');
17 c = getchar();
18 }
19 return f ? -a : a;
20 }
21
22 int pot[15010] , num[40010] , A[15010] , B[15010] , C[15010] , D[15010];
23
24 int main(){
25 #ifndef ONLINE_JUDGE
26 freopen("2119.in" , "r" , stdin);
27 //freopen("2119.out" , "w" , stdout);
28 #endif
29 int N = read() , M = read();
30 for(int i = 1 ; i <= M ; i++)
31 ++pot[num[i] = read()];
32 for(int i = 1 ; i * 9 < N ; i++){
33 int sum = 0;
34 for(int d = 9 * i + 2 ; d <= N ; d++){
35 int c = d - i;
36 sum += pot[d - 7 * i - 1] * pot[d - 9 * i - 1];
37 C[c] += sum * pot[d];
38 D[d] += sum * pot[c];
39 }
40 sum = 0;
41 for(int a = N - 9 * i - 1 ; a ; a--){
42 int b = a + 2 * i;
43 sum += pot[b + 6 * i + 1] * pot[b + 7 * i + 1];
44 B[b] += sum * pot[a];
45 A[a] += sum * pot[b];
46 }
47 }
48 for(int i = 1 ; i <= M ; i++)
49 printf("%d %d %d %d\n" , A[num[i]] , B[num[i]] , C[num[i]] , D[num[i]]);
50 return 0;
51 }
NOIP2017(clear)
1 #include
2 using namespace std;
3 int main(){
4 int a , b , c;
5 cin >> a >> b >> c;
6 cout << a / 5 + b / 10 * 3 + c / 2;
7 return 0;
8 }
1 #include
2 using namespace std;
3 int book[1001];
4 int main()
5 {
6 int n , q;
7 cin >> n >> q;
8 for(int i = 1 ; i <= n ; i++)
9 cin >> book[i];
10 sort(book + 1 , book + n + 1);
11 while(q--){
12 int l , a , i;
13 cin >> l >> a;
14 l = pow(10 , l);
15 for(i = 1 ; i <= n ; i++)
16 if(book[i] % l == a)
17 break;
18 if(i == n + 1)
19 cout << -1 << endl;
20 else
21 cout << book[i] << endl;
22 }
23 return 0;
24 }
1 //瞎搞图论
2 #include
3 #define inf 0x3f3f3f3f
4 #define X now.x + dir[i][0]
5 #define Y now.y + dir[i][1]
6 //This code is written by Itst
7 using namespace std;
8
9 inline int read(){
10 int a = 0;
11 bool f = 0;
12 char c = getchar();
13 while(c != EOF && !isdigit(c)){
14 if(c == '-')
15 f = 1;
16 c = getchar();
17 }
18 while(c != EOF && isdigit(c)){
19 a = (a << 3) + (a << 1) + (c ^ '0');
20 c = getchar();
21 }
22 return f ? -a : a;
23 }
24
25 struct node{
26 int x , y , col;
27 }now;
28 const int dir[4][2] = {0 , 1 , 0 , -1 , 1 , 0 , -1 , 0};
29 int minDis[102][102][3] , col[102][102] , M , N;
30 queue < node > q;
31
32 void SPFA(){
33 memset(minDis , 0x3f , sizeof(minDis));
34 minDis[1][1][col[1][1]] = 0;
35 q.push((node){1 , 1 , col[1][1]});
36 while(!q.empty()){
37 now = q.front();
38 q.pop();
39 for(int i = 0 ; i < 4 ; i++)
40 if(col[X][Y] + 1)
41 if(col[X][Y] || col[now.x][now.y]){
42 if(col[X][Y]){
43 if(minDis[X][Y][col[X][Y]] > minDis[now.x][now.y][now.col] + (now.col != col[X][Y])){
44 minDis[X][Y][col[X][Y]] = minDis[now.x][now.y][now.col] + (now.col != col[X][Y]);
45 q.push((node){X , Y , col[X][Y]});
46 }
47 }
48 else{
49 col[X][Y] = now.col;
50 if(minDis[X][Y][col[X][Y]] > minDis[now.x][now.y][now.col] + 2){
51 minDis[X][Y][col[X][Y]] = minDis[now.x][now.y][now.col] + 2;
52 q.push((node){X , Y , col[X][Y]});
53 }
54 col[X][Y] = 0;
55 }
56 }
57 }
58 }
59
60 int main(){
61 #ifndef ONLINE_JUDGE
62 freopen("3956.in" , "r" , stdin);
63 //freopen("3956.out" , "w" , stdout);
64 #endif
65 memset(col , -1 , sizeof(col));
66 M = read();
67 N = read();
68 for(int i = 1 ; i <= M ; i++)
69 for(int j = 1 ; j <= M ; j++)
70 col[i][j] = 0;
71 while(N--){
72 int a = read() , b = read() , c = read() + 1;
73 col[a][b] = c;
74 }
75 SPFA();
76 if(minDis[M][M][0] == inf && minDis[M][M][1] == inf && minDis[M][M][2] == inf)
77 puts("-1");
78 else
79 printf("%d\n" , min(minDis[M][M][0] , min(minDis[M][M][1] , minDis[M][M][2])));
80 return 0;
81 }
1 //二分答案、单调队列优化DP
2 #include
3 #define MAXN 500002
4 using namespace std;
5 inline int read(){
6 int a = 0;
7 bool f = 0;
8 char c = getchar();
9 while(!isdigit(c)){
10 if(c == '-') f = 1;
11 c = getchar();
12 }
13 while(isdigit(c)) a += (a << 3) + a + (c ^ '0') , c = getchar();
14 return f ? -a : a;
15 }
16 inline int max(int a , int b){return a > b ? a : b;}
17 int num[MAXN][2] , maxN[MAXN] , N , d , K;
18 bool vis[MAXN];
19 bool check(int mid){
20 deque < int > q;
21 int ALLmax = 0 , r = 0;
22 for(int i = 1 ; i <= N && ALLmax < K ; i++){
23 while(num[i][0] - num[r][0] > d + mid) r++;
24 while(num[i][0] - num[r][0] >= max(d - mid , 1) && num[i][0] - num[r][0] <= d + mid){
25 if(vis[r]){
26 while(!q.empty() && maxN[q.back()] <= maxN[r])
27 q.pop_back();
28 q.push_back(r);
29 }
30 r++;
31 }
32 while(!q.empty() && num[i][0] - num[q.front()][0] > d + mid)
33 q.pop_front();
34 if(!q.empty()){
35 ALLmax = max(ALLmax , maxN[i] = maxN[q.front()] + num[i][1]);
36 vis[i] = 1;
37 }
38 else vis[i] = 0;
39 }
40 return ALLmax >= K;
41 }
42 int main(){
43 N = read(); d = read(); K = read();
44 for(int i = 1 ; i <= N ; i++){
45 num[i][0] = read();
46 num[i][1] = read();
47 }
48 vis[0] = 1;
49 int l = 0 , r = 1e9 + 1;
50 while(l < r){
51 int mid = l + r >> 1;
52 if(check(mid)) r = mid;
53 else l = mid + 1;
54 }
55 if(r == 1e9 + 1) cout << -1;
56 else cout << l;
57 return 0;
58 }