2017中国大学生程序设计竞赛 - 女生专场

比赛链接：http://acm.hdu.edu.cn/contests/contest_show.php?cid=772

昨天嘴巴了5题，结果今天错了2个。真弱啊。。

 

1001.水

 1 #include 
 2 using namespace std;
 3 
 4 const int maxn = 15;
 5 int n, m;
 6 int vis[maxn];
 7 int solve[maxn];
 8 char st[22];
 9 
10 int main() {
11     // freopen("in", "r", stdin);
12     int T;
13     scanf("%d", &T);
14     while(T--) {
15         scanf("%d%d", &n,&m);
16         int id, a, b;
17         memset(vis, 0, sizeof(vis));
18         memset(solve, 0, sizeof(solve));
19         for(int i = 0; i < m; i++) {
20             scanf("%d", &id);
21             id -= 1000;
22             scanf("%d:%d",&a,&b);
23             scanf("%s", st);
24             if(solve[id]) continue;
25             if(st[0] == 'A') {
26                 solve[id] = 1;
27                 vis[id] += a * 60 + b;
28             } 
29             else vis[id] += 20;
30         }
31         int tot = 0, ret = 0;
32         for(int i = 1; i <= n; i++) {
33             if(solve[i]) {
34                 tot++;
35                 ret += vis[i];
36             }
37         }
38         printf("%d %d\n", tot, ret);
39     }
40     return 0;
41 }

 

1002. f(i,j)代表1~i个教室，并且在i上建了一共j个shop的最小花费，更新从f(i-1,j)更新来，在j!=i的时候则是i到x(j)的花费+f(i-1,j)的花费和，最后记录下最小的花费，更新在i点建shop的时候的花费。

小心给的数据可能不是坐标递增的，所以排个序。

 1 #include 
 2 using namespace std;
 3 
 4 #define x first
 5 #define c second
 6 typedef long long LL;
 7 typedef pair pll;
 8 const int maxn = 3030;
 9 const LL inf = 1LL << 61;
10 int n;
11 pll p[maxn];
12 LL f[maxn][maxn];
13 
14 int main() {
15     // freopen("in", "r", stdin);
16     while(~scanf("%d", &n)) {
17         memset(f, 0, sizeof(f));
18         for(int i = 1; i <= n; i++) {
19             scanf("%I64d%I64d",&p[i].x,&p[i].c);
20         }
21         sort(p+1, p+n+1);
22         f[1][1] = p[1].c;
23         for(int i = 2; i <= n; i++) {
24             LL cur = f[i-1][1];
25             for(int j = 1; j < i; j++) {
26                 cur = min(cur, f[i-1][j]);
27                 f[i][j] = f[i-1][j] + p[i].x - p[j].x;
28             }
29             f[i][i] = cur + p[i].c;
30         }
31         LL ret = f[n][1];
32         for(int i = 2; i <= n; i++) {
33             ret = min(ret, f[n][i]);
34         }
35         cout << ret << endl;
36     }
37     return 0;
38 }

 

1003.维护前缀gcd和后缀gcd，删的时候求某点两侧gcd的gcd就行了。

 1 #include 
 2 using namespace std;
 3 
 4 const int maxn = 100100;
 5 int n, ret;
 6 int a[maxn];
 7 int f[3][maxn];
 8 int gcd(int x, int y) {
 9     return y == 0 ? x : gcd(y, x%y);
10 }
11 
12 int main() {
13     // freopen("in", "r", stdin);
14     int T;
15     scanf("%d", &T);
16     while(T--) {
17         scanf("%d", &n);
18         for(int i = 1; i <= n; i++) {
19             scanf("%d", &a[i]);
20         }
21         ret = 0;
22         f[0][1] = a[1]; f[1][n] = a[n];
23         for(int i = 2; i <= n; i++) f[0][i] = gcd(f[0][i-1], a[i]);
24         for(int i = n - 1; i >= 1; i--) f[1][i] = gcd(f[1][i+1], a[i]);
25         ret = max(f[1][2], f[0][n-1]);
26         for(int i = 2; i <= n - 1; i++) {
27             ret = max(ret, gcd(f[0][i-1], f[1][i+1]));
28         }
29         printf("%d\n", ret);
30     }
31     return 0;
32 }

 

1004.相当于问删掉一条一条边以后，仍然能构成原图一样的最小树形图，问一共可以有多少种删边情况。先跑最短路，选中某点的属于最短路上的入边，将这些入边组合起来便是最小树形图。相当于问有多少种入边权值相同但是边不同的情况。统计某点所有和最短路长度相同的边的个数作为贡献，最后将所有点的贡献乘起来就行。注意有不连通的情况。

 1 #include 
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 typedef pairint> pli;
 6 const LL mod = (LL)1e9+7;
 7 const LL inf = 1LL << 62;
 8 const int maxn = 55;
 9 priority_queue, greater > pq;
10 int n;
11 char G[maxn][maxn];
12 bool vis[maxn];
13 int to[maxn][maxn];
14 LL d[maxn];
15 LL mut[maxn];
16 
17 int dij(int s) {
18     for(int i = 0; i < n; i++) d[i] = inf;
19     memset(vis, 0, sizeof(vis));
20     while(!pq.empty()) pq.pop();
21     vis[s] = 1; d[s] = 0;
22     pq.push(pli(0, s));
23     while(!pq.empty()) {
24         pli tmp = pq.top(); pq.pop();
25         LL pw = tmp.first;
26         int u = tmp.second;
27         for(int v = 0; v < n; v++) {
28             if(vis[v]) continue;
29             if(G[u][v] == '0') continue;
30             if(d[v] > d[u] + G[u][v] - '0') {
31                 d[v] = d[u] + G[u][v] - '0';
32                 pq.push(pli(d[v], v));
33             }
34         }
35     }
36     for(int i = 0; i < n; i++) if(d[i] == inf) return 0;
37     return 1;
38 }
39 
40 void dfs(int u) {
41     for(int v = 0; v < n; v++) {
42         if(G[u][v]-'0' && d[v] == d[u]+G[u][v]-'0') {
43             if(to[u][v]) continue;
44             to[u][v] = 1; mut[v]++;
45             dfs(v);
46         }
47     }
48 }
49 
50 int main() {
51     // freopen("in", "r", stdin);
52     while(~scanf("%d", &n)) {
53         memset(mut, 0, sizeof(mut));
54         memset(to, 0, sizeof(to));
55         for(int i = 0; i < n; i++) {
56             scanf("%s", G[i]);
57         }
58         int ok = dij(0);
59         dfs(0);
60         LL ret = 1;
61         for(int i = 1; i < n; i++) ret = ret * mut[i] % mod;
62         if(!ok) puts("0");
63         else printf("%I64d\n", ret);
64     }
65     return 0;
66 }

 

1005.快速幂水过。

 1 #include 
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 const LL mod = (LL)1e9+7;
 6 LL n, k;
 7 
 8 LL mul(LL i, LL k) {
 9     LL ret = 1;
10     while(k) {
11         if(k & 1) ret = (ret * i) % mod;
12         i = (i * i) % mod;
13         k >>= 1;
14     }
15     return ret;
16 }
17 
18 int main() {
19     // freopen("in", "r", stdin);
20     int T;
21     scanf("%d", &T);
22     while(T--) {
23         scanf("%I64d%I64d",&n,&k);
24         LL ret = 0;
25         for(LL i = 1; i <= n; i++) {
26             ret = (ret + mul(i, k)) % mod;
27         }
28         cout << ret % mod << endl;
29     }
30     return 0;
31 }

 

1007.贪心，从左往右扫，遇到2则记录值+1，希望让后面的1与它匹配。假如没有2的边却出现了1的边，则暂时不计数。最后看计数结果是不是0。注意奇数点直接输出No。

 1 #include 
 2 using namespace std;
 3 
 4 const int maxn = 100100;
 5 int n;
 6 
 7 int main() {
 8     // freopen("in", "r", stdin);
 9     int T, v;
10     scanf("%d", &T);
11     while(T--) {
12         scanf("%d", &n);
13         int two = 0;
14         for(int i = 2; i <= n; i++) {
15             scanf("%d", &v);
16             if(v == 1) {
17                 if(two) two--;
18             }
19             else two++;
20         }
21         if(n & 1) {
22             printf("No\n");
23             continue;
24         }
25         if(!two) printf("Yes\n");
26         else printf("No\n");
27     }
28     return 0;
29 }

 

1008.打表发现递推式f(n)=f(n-1)+f(n-3)，f(2)=3，f(3)=4，f(4)=6。

构造矩阵：

1 0 1

1 0 0

0 1 0

快速水过。

 1 #include 
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 
 6 const LL mod = (LL)1E9+7;
 7 const LL maxn = 11;
 8 typedef struct Matrix {
 9     LL m[maxn][maxn];
10     LL r;
11     LL c;
12     Matrix() {
13         r = c = 0;
14         memset(m, 0, sizeof(m));
15     } 
16 } Matrix;
17 Matrix mul(Matrix m1, Matrix m2) {
18     Matrix ans = Matrix();
19     ans.r = m1.r; ans.c = m2.c;
20     for(LL i = 1; i <= m1.r; i++) {
21         for(LL j = 1; j <= m2.r; j++) {
22             for(LL k = 1; k <= m2.c; k++) {
23                 if(m2.m[j][k] == 0) continue;
24                 ans.m[i][k] = ((ans.m[i][k] + (m1.m[i][j] * m2.m[j][k]) % mod) % mod) % mod;
25             }
26         }
27     }
28     return ans;
29 }
30 Matrix quickmul(Matrix m, LL n) {
31     Matrix ans = Matrix();
32     for(LL i = 1; i <= m.r; i++) ans.m[i][i] = 1;
33     ans.r = m.r;
34     ans.c = m.c;
35     while(n) {
36         if(n & 1) ans = mul(m, ans);
37         m = mul(m, m); n >>= 1;
38     }
39     return ans;
40 }
41 
42 LL n;
43 
44 inline bool scan_d(LL &num) {
45     char in;bool IsN=false;
46     in=getchar();
47     if(in==EOF) return false;
48     while(in!='-'&&(in<'0'||in>'9')) in=getchar();
49     if(in=='-'){ IsN=true;num=0;}
50     else num=in-'0';
51     while(in=getchar(),in>='0'&&in<='9'){
52         num*=10,num+=in-'0';
53     }
54     if(IsN) num=-num;
55     return true;
56 }
57 
58 
59 int main() {
60     // freopen("in", "r", stdin);
61     LL T;
62     scan_d(T);
63     while(T--) {
64         scan_d(n);
65         if(n == 2) {
66             printf("3\n");
67             continue;
68         }
69         if(n == 3) {
70             printf("4\n");
71             continue;
72         }
73         if(n == 4) {
74             printf("6\n");
75             continue;
76         }
77         Matrix a; a.r = a.c = 3;
78         a.m[1][1] = 1; a.m[1][2] = 0; a.m[1][3] = 1;
79         a.m[2][1] = 1; a.m[2][2] = 0; a.m[2][3] = 0;
80         a.m[3][1] = 0; a.m[3][2] = 1; a.m[3][3] = 0;
81         Matrix b = quickmul(a, n-4);
82         Matrix p; p.r = 3; p.c = 1;
83         p.m[1][1] = 6; p.m[2][1] = 4; p.m[3][1] = 3;
84         p = mul(b, p);
85         printf("%I64d\n", (p.m[1][1]) % mod);
86     }
87     return 0;
88 }

 

转载于:https://www.cnblogs.com/kirai/p/6821524.html