June Challenge 2018 Division 2

Naive Chef

暴力

 1 #include 
 2 using namespace std;
 3  
 4 int main() {
 5     int T;
 6     scanf("%d", &T);
 7     while(T--){
 8         int N, A, B, x, ca = 0, cb = 0;
 9         scanf("%d %d %d", &N, &A, &B);
10         for(int i = 1; i <= N; ++i) {
11             scanf("%d", &x);
12             if(x == A) ca++;
13             if(x == B) cb++;
14         }
15         printf("%f\n", 1.0 * ca * cb / N / N);
16     }
17     return 0;
18 } 

Aguin

 

Binary Shuffle

每次可以加一个或者减到剩一个 特判几个点

 1 #include 
 2 using namespace std;
 3 typedef long long LL;
 4  
 5 int main() {
 6     int T;
 7     scanf("%d", &T);
 8     while (T--) {
 9         LL A, B, ca = 0, cb = 0;
10         scanf("%lld %lld", &A, &B);
11         if(A == B) {puts("0"); continue;}
12         if(B == 0) {puts("-1"); continue;}
13         B--;
14         for (int i = 0; i <= 62; ++i) {
15             if ((1LL << i) & A) ca++;
16             if ((1LL << i) & B) cb++;
17         }
18         if(B == 0) {puts(A == 0 ? "1" : "-1"); continue;}
19         if(ca == cb) puts("1");
20         else if(ca > cb) puts("2");
21         else printf("%d\n", cb - ca + 1);
22     }
23     return 0;
24 } 

Aguin

 

Vision

二分 我连三维叉积都不会了

 1 using namespace std;
 2 const double eps = 1e-9;
 3  
 4 double sqr(double x) {
      return x * x;}
 5 double dis(double x1, double y1, double z1, double x2, double y2, double z2) {
 6     return sqrt(sqr(x1 - x2) + sqr(y1 - y2) + sqr(z1 - z2));
 7 }
 8 double cross(double x1, double y1, double z1, double x2, double y2, double z2) {
 9     return sqrt(sqr(y1 * z2 - z1 * y2) + sqr(z1 * x2 - x1 * z2) + sqr(x1 * y2 - y1 * x2));
10 }
11  
12 int main() {
13     int T;
14     scanf("%d", &T);
15     while(T--) {
16         double Px, Py, Pz;
17         scanf("%lf %lf %lf", &Px, &Py, &Pz);
18         double Qx, Qy, Qz;
19         scanf("%lf %lf %lf", &Qx, &Qy, &Qz);
20         double dx, dy, dz;
21         scanf("%lf %lf %lf", &dx, &dy, &dz);
22         double cx, cy, cz, r;
23         scanf("%lf %lf %lf %lf", &cx, &cy, &cz, &r);
24         double ans = 1e10, tmp = 1e10;
25         while(tmp > eps) {
26             double M = ans - tmp;
27             double xx = Qx + dx * M;
28             double yy = Qy + dy * M;
29             double zz = Qz + dz * M;
30             double PQ = dis(Px, Py, Pz, xx, yy, zz);
31             double d = fabs(cross(Px - cx, Py - cy, Pz - cz, xx - cx, yy - cy, zz - cz)) / PQ;
32             if(d >= r) ans = M;
33             tmp /= 2;
34         }
35         printf("%.8f\n", ans);
36     }
37     return 0;
38 } 

Aguin

 

Sheokand and String

字典树

 1 #include 
 2 using namespace std;
 3 const int maxn = 1e6 + 10;
 4 string S[maxn], P[maxn], ans[maxn];
 5 int R[maxn], id[maxn];
 6 char str[22];
 7  
 8 bool cmp(int i, int j) {
 9     return R[i] < R[j];
10 }
11  
12 int cnt, nxt[maxn][26], val[maxn];
13 void INS(int x) {
14     int u = 0;
15     for (int i = 0; i < S[x].length(); ++i) {
16         if (nxt[u][S[x][i] - 'a']) u = nxt[u][S[x][i] - 'a'];
17         else u = nxt[u][S[x][i] - 'a'] = ++cnt;
18     }
19     val[u] = 1;
20 }
21 string GET(int x) {
22     string ret = "";
23     int u = 0, flag = 0;
24     for (int i = 0; i < P[x].length(); ++i) {
25         if (!flag && nxt[u][P[x][i] - 'a']) ret += P[x][i], u = nxt[u][P[x][i] - 'a'];
26         else {
27             if (val[u]) return ret;
28             flag = 1;
29         }
30         if (flag)
31             for (int j = 0; j < 26; ++j) {
32                 if (nxt[u][j]) {
33                     u = nxt[u][j];
34                     ret += 'a' + j;
35                     if (val[u]) return ret;
36                     break;
37                 }
38             }
39     }
40     if (val[u]) return ret;
41     while (1) {
42         for (int j = 0; j < 26; ++j) {
43             if (nxt[u][j]) {
44                 u = nxt[u][j];
45                 ret += 'a' + j;
46                 if (val[u]) return ret;
47                 break;
48             }
49         }
50     }
51 }
52  
53 int main(){
54     int N, Q;
55     scanf("%d", &N);
56     for(int i = 1; i <= N; ++i){
57         scanf("%s", str);
58         S[i] = string(str);
59     }
60     scanf("%d", &Q);
61     for(int i = 1; i <= Q; ++i){
62         scanf("%d %s", R + i, str);
63         P[i] = string(str);
64         id[i] = i;
65     }
66     sort(id + 1, id + 1 + Q, cmp);
67     int p = 0;
68     for(int i = 1; i <= Q; ++i) {
69         int x = id[i];
70         while(p + 1 <= R[x]) INS(++p);
71         ans[x] = GET(x);
72     }
73     for(int i = 1; i <= Q; ++i) printf("%s\n", ans[i].c_str());
74     return 0;
75 } 

Aguin

 

Two Flowers

枚举一个颜色的块 然后用并查集维护这个颜色以外的连通性 每条相邻边合并一次 所以是On的

 1 #include 
 2 using namespace std;
 3 int ans, a[2222][2222], id[2222][2222];
 4  
 5 typedef pair<int, int> pii;
 6 map< int, vector > M;
 7 map< int, vector > :: iterator it;
 8  
 9 int r[4444444];
10 int fa[4444444];
11 stack FA, R;
12 int Find(int x) {
13     return x == fa[x] ? x : Find(fa[x]);
14 }
15 void Union(int x, int y) {
16     x = Find(x), y = Find(y);
17     if(x == y) return;
18     if(r[x] < r[y]) swap(x, y);
19     FA.push(pii(y, fa[y]));
20     R.push(pii(x, r[x]));
21     fa[y] = x, r[x] += r[y];
22 }
23  
24 int dx[] = {
      1, 0, -1, 0};
25 int dy[] = {
      0, 1, 0, -1};
26 void dfs(int i, int j, int x) {
27     id[i][j] = x;
28     r[x]++;
29     ans = max(ans, r[x]);
30     for (int o = 0; o < 4; ++o) {
31         int nx = i + dx[o], ny = j + dy[o];
32         if (a[nx][ny] == a[i][j] && !id[nx][ny]) dfs(nx, ny, x);
33     }
34 }
35  
36 int cnt, cnt2, vis[2222][2222];
37 map<int, int> mp;
38 void dfs1(int i, int j) {
39     vis[i][j] = 1;
40     for (int o = 0; o < 4; ++o) {
41         int nx = i + dx[o], ny = j + dy[o];
42         if (a[nx][ny] && a[nx][ny] != a[i][j]) {
43             if (mp.find(a[nx][ny]) == mp.end())
44                 mp[a[nx][ny]] = ++cnt2 + cnt, fa[cnt + cnt2] = cnt + cnt2, r[cnt + cnt2] = r[id[i][j]];
45             Union(id[nx][ny], mp[a[nx][ny]]);
46             ans = max(ans, r[Find(id[nx][ny])]);
47         }
48         if (a[nx][ny] == a[i][j] && !vis[nx][ny]) dfs1(nx, ny);
49     }
50 }
51  
52 void cln(){
53     while(!FA.empty()) {
54         pii x = FA.top(); FA.pop();
55         fa[x.first] = x.second;
56     }
57     while(!R.empty()) {
58         pii x = R.top(); R.pop();
59         r[x.first] = x.second;
60     }
61 }
62  
63 int main() {
64     int n, m;
65     scanf("%d %d", &n, &m);
66     for(int i = 1; i <= n; ++i)
67         for(int j = 1; j <= m; ++j)
68             scanf("%d", &a[i][j]);
69     for(int i = 1; i <= n; ++i) {
70         for(int j = 1; j <= m; ++j) {
71             if(id[i][j]) continue;
72             dfs(i, j, ++cnt), fa[cnt] = cnt;
73             M[a[i][j]].push_back(pii(i, j));
74         }
75     }
76     for(it = M.begin(); it != M.end(); it++) {
77         cnt2 = 0;
78         vector & v = (*it).second;
79         for(int i = 0; i < v.size(); ++i) {
80             int x = v[i].first, y = v[i].second;
81             mp.clear(), dfs1(x, y);
82         }
83         cln();
84     }
85     printf("%d\n", ans);
86     return 0;
87 } 

Aguin

 

Ways to Work

考虑$\{d_i-i+1\}$这个序列，每次可以加任意非负整数或者减一，枚举C的因子作为$d_{n}$然后倒着往前贪心，如果能+1就+1，否则减到最大的能整除的因子

 1 #include 
 2 using namespace std;
 3 const int maxn = 1e6 + 10;
 4 vector<int> fac, ans;
 5  
 6 int main() {
 7     int T;
 8     scanf("%d", &T);
 9     while(T--) {
10         int N, C;
11         scanf("%d %d", &N, &C);
12         fac.clear();
13         for(int i = 1; i <= C / i; ++i) {
14             if(C % i == 0) {
15                 fac.push_back(i);
16                 if(i != C / i) fac.push_back(C / i);
17             }
18         }
19         sort(fac.begin(), fac.end());
20         for(int i = 0; i < fac.size(); ++i) {
21             ans.clear();
22             int cnt = 0, tmp = C;
23             for(int j = i; ; ) {
24                 while (tmp % fac[j] != 0) j--;
25                 tmp /= fac[j], cnt++, ans.push_back(fac[j]);
26                 if(cnt == N || tmp == 1) break;
27                 if(tmp % (fac[j] + 1) == 0) j++;
28             }
29             if(tmp == 1) break;
30         }
31         for(int i = ans.size() + 1; i <= N; ++i) ans.push_back(1);
32         for(int i = 0; i < N; ++i) printf("%d%c", ans[N-1-i] + i, i == N - 1 ? '\n' : ' ');
33     }
34     return 0;
35 } 

Aguin

 

Expected Buildings

只要求N段a的和，利用矩阵快速幂求前缀和减一下就可以了，又是向量乘……

 1 #include 
 2 using namespace std;
 3 typedef long long LL;
 4 const LL mod = 163577857;
 5 int p[50005], a[105], c[105];
 6 int N, h, x, K;
 7  
 8 struct Matrix {
 9     LL a[105][105];
10     void init() {
11         memset(a, 0, sizeof(a));
12         for (int i = 0; i < 105; ++i) {
13             a[i][i] = 1;
14         }
15     }
16 } P[30];
17  
18 Matrix mul(Matrix a, Matrix b) {
19     Matrix ans;
20     memset(ans.a, 0, sizeof(ans.a));
21     for (int i = 0; i < 105; ++i) {
22         for (int k = 0; k < 105; ++k) {
23             if (a.a[i][k] != 0)
24                 for (int j = 0; j < 105; ++j) {
25                     ans.a[i][j] = (ans.a[i][j] + a.a[i][k] * b.a[k][j]) % mod;
26                 }
27         }
28     }
29     return ans;
30 }
31  
32 LL sum[105], b[105], cpy[105];
33 LL cal(int o) {
34     if(o <= K) return sum[o];
35     o -= K;
36     for(int i = 1; i <= K; ++i) b[i] = a[1+K-i];
37     b[K+1] = sum[K];
38     for(int i = 0; i < 30; ++i) {
39         if(o & (1 << i)) {
40             memset(cpy, 0, sizeof(cpy));
41             for(int j = 1; j <= K + 1; ++j)
42                 for(int k = 1; k <= K + 1; ++k)
43                     cpy[j] = (cpy[j] + P[i].a[j][k] * b[k]) % mod;
44             memcpy(b, cpy, sizeof(b));
45         }
46     }
47     return b[K+1];
48 }
49  
50 LL fp(LL a, int b) {
51     LL ret = 1;
52     while (b) {
53         if (b & 1) ret = ret * a % mod;
54         a = a * a % mod;
55         b >>= 1;
56     }
57     return ret;
58 }
59  
60 int main() {
61     scanf("%d %d %d %d", &N, &h, &x, &K);
62     for(int i = 1; i <= N; ++i) scanf("%d", p + i);
63     for(int i = 1; i <= K; ++i) scanf("%d", a + i), sum[i] = (sum[i-1] + a[i]) % mod;
64     for(int i = 1; i <= K; ++i) scanf("%d", c + i);
65     for(int i = 1; i <= K; ++i) P[0].a[1][i] = c[i];
66     for(int i = 2; i <= K; ++i) P[0].a[i][i-1] = 1;
67     for(int i = 1; i <= K; ++i) P[0].a[K+1][i] = c[i];
68     P[0].a[K+1][K+1] = 1;
69     for(int i = 1; i < 30; ++i) P[i] = mul(P[i-1], P[i-1]);
70     LL ans = 0;
71     for(int i = 1; i <= N; ++i) {
72         if(p[i] < x) ans = (ans + cal(p[i]) + cal(h) - cal(h - x + p[i]) + mod) % mod;
73         else ans = (ans + cal(p[i]) - cal(p[i] - x) + mod) % mod;
74     }
75     printf("%lld\n", ans * fp(cal(h), mod - 2) % mod);
76     return 0;
77 } 

Aguin

 

Warehouseman (Challenge)

转载于:https://www.cnblogs.com/Aguin/p/9175991.html