暴力
1 #include <bits/stdc++.h> 2 using namespace std; 3 char s[111]; 4 5 int main() { 6 scanf("%s", s + 1); 7 int l = strlen(s + 1), ok = 0; 8 for(int i = 1; i <= l; ++i) { 9 if(s[i] == 'A' && s[i-1] == 'B' && s[i+1] == 'C') ok = 1; 10 if(s[i] == 'A' && s[i-1] == 'C' && s[i+1] == 'B') ok = 1; 11 if(s[i] == 'B' && s[i-1] == 'A' && s[i+1] == 'C') ok = 1; 12 if(s[i] == 'B' && s[i-1] == 'C' && s[i+1] == 'A') ok = 1; 13 if(s[i] == 'C' && s[i-1] == 'A' && s[i+1] == 'B') ok = 1; 14 if(s[i] == 'C' && s[i-1] == 'B' && s[i+1] == 'A') ok = 1; 15 } 16 puts(ok ? "Yes" : "No"); 17 return 0; 18 }
暴力（这不本来就On吗- -）
1 #include <bits/stdc++.h> 2 using namespace std; 3 char s[2222]; 4 5 int main() { 6 int n, p, ok = 0; 7 scanf("%d %d %s", &n, &p, s + 1); 8 for(int i = p + 1; i <= n; ++i) { 9 if(s[i] == '.') { 10 if(s[i-p] == '.') s[i-p] = '0', s[i] = '1', ok = 1; 11 else s[i] = '1' + '0' - s[i-p], ok = 1; 12 } 13 else { 14 if(s[i-p] == '.') s[i-p] = '1' + '0' - s[i], ok = 1; 15 else if(s[i-p] != s[i]) ok = 1; 16 } 17 } 18 if(!ok) puts("No"); 19 else { 20 for(int i = 1; i <= n; ++i) 21 if(s[i] == '.') s[i] = '0'; 22 puts(s + 1); 23 } 24 return 0; 25 }
大概的想法都是把一个颜色嵌到另一个颜色里面
1 #include <bits/stdc++.h> 2 using namespace std; 3 int G[55][55]; 4 5 int main() { 6 int a, b, c, d; 7 scanf("%d %d %d %d", &a, &b, &c, &d); 8 a--, b--; 9 for(int i = 1; i <= 50; ++i) { 10 for(int j = 1; j <= 25; ++j) { 11 if(i & 1) G[i][j] = 1; 12 else if(j & 1) G[i][j] = 1; 13 } 14 for(int j = 26; j <= 50; ++j){ 15 if(i & 1) G[i][j] = 2; 16 else if(!(j & 1)) G[i][j] = 2; 17 } 18 } 19 for(int i = 1; i <= 50; ++i) 20 for(int j = 1; j <= 25; ++j) 21 if(b && !G[i][j]) b--, G[i][j] = 2; 22 for(int i = 1; i <= 50; ++i) 23 for(int j = 26; j <= 50; ++j) 24 if(a && !G[i][j]) a--, G[i][j] = 1; 25 for(int i = 1; i <= 50; ++i) 26 for(int j = 1; j <= 50; ++j) 27 if(c && !G[i][j]) c--, G[i][j] = 3; 28 for(int i = 1; i <= 50; ++i) 29 for(int j = 1; j <= 50; ++j) 30 if(d && !G[i][j]) d--, G[i][j] = 4; 31 for(int i = 1; i <= 50; ++i) 32 for(int j = 1; j <= 25; ++j) 33 if(!G[i][j]) G[i][j] = 1; 34 for(int i = 1; i <= 50; ++i) 35 for(int j = 26; j <= 50; ++j) 36 if(!G[i][j]) G[i][j] = 2; 37 puts("50 50"); 38 for(int i = 1; i <= 50; ++i){ 39 for(int j = 1; j <= 50; ++j) { 40 putchar('A' - 1 + G[i][j]); 41 } 42 puts(""); 43 } 44 return 0; 45 }
转化成月亮动比较显然，然后利用x-t图写出满足要求的不等式，枚举一个方向，另一个方向解不等式二分
我特判了一个w=1的情况
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 vector<LL> L, R; 5 6 int main() { 7 LL n, l, w; 8 cin >> n >> l >> w; 9 for(int i = 1; i <= n; ++i) { 10 LL x, v; 11 cin >> x >> v; 12 if(v == -1) L.push_back(x); 13 else R.push_back(x); 14 } 15 sort(R.begin(), R.end()); 16 LL ans = 0; 17 for(int i = 0; i < L.size(); ++i) { 18 if(w == 1 && L[i] + l <= 0) continue; 19 int lb = lower_bound(R.begin(), R.end(), 1.0 * (w - 1) * (L[i] + l) / (w + 1)) - R.begin(); 20 if(w != 1) lb = min(lb, lower_bound(R.begin(), R.end(), 1.0 * (w + 1) * (L[i] + l) / (w - 1)) - R.begin()); 21 ans += lb; 22 } 23 cout << ans << endl; 24 return 0; 25 }
很显然的矩阵快速幂
对于第一步，一定是选一条线而不是线的交点，因为平均不会更好
又是要把矩阵乘转变为向量乘来优化掉一个n的无趣写法
1 #include <bits/stdc++.h> 2 using namespace std; 3 const double pi = acos(-1); 4 const double eps = 1e-8; 5 const int maxn = 205; 6 int x[maxn], y[maxn]; 7 8 struct Matrix { 9 double a[maxn][maxn]; 10 void init() { 11 memset(a, 0, sizeof(a)); 12 for (int i = 0; i < maxn; ++i) { 13 a[i][i] = 1; 14 } 15 } 16 } P[14]; 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 < maxn; ++i) { 22 for (int k = 0; k < maxn; ++k) { 23 if (a.a[i][k] != 0) 24 for (int j = 0; j < maxn; ++j) { 25 ans.a[i][j] += a.a[i][k] * b.a[k][j]; 26 } 27 } 28 } 29 return ans; 30 } 31 32 typedef pair<double, int> pdi; 33 vector<pdi> v; 34 35 double d[maxn], e[maxn]; 36 int main() { 37 int n, q; 38 scanf("%d", &n); 39 for (int i = 1; i <= n; ++i) scanf("%d %d", x + i, y + i); 40 for (int i = 1; i <= n; ++i) { 41 v.clear(); 42 for (int j = 1; j <= n; ++j) { 43 if (j == i) continue; 44 double theta = atan2(x[j] - x[i], y[j] - y[i]); 45 while (theta < 0) theta += pi; 46 while (theta >= pi) theta -= pi; 47 v.push_back(pdi(theta, j)); 48 } 49 sort(v.begin(), v.end()); 50 int line = 0; 51 for (int j = 0, k; j < n - 1;) { 52 for (k = j; k < n - 1 && fabs(v[k].first - v[j].first) < eps; ++k); 53 j = k, line++; 54 } 55 for (int j = 0, k; j < n - 1;) { 56 for (k = j; k < n - 1 && fabs(v[k].first - v[j].first) < eps; ++k); 57 for (int p = j; p < k; ++p) P[0].a[v[p].second][i] += 1.0 / line / (k - j + 1); 58 P[0].a[i][i] += 1.0 / line / (k - j + 1); 59 j = k; 60 } 61 } 62 for(int i = 1; i < 14; ++i) P[i] = mul(P[i-1], P[i-1]); 63 scanf("%d", &q); 64 while (q--) { 65 int t, m; 66 scanf("%d %d", &t, &m); 67 m--; 68 memset(d, 0, sizeof(d)); 69 d[t] = 1; 70 for(int i = 0; i < 14; ++i) { 71 if(m & (1 << i)) { 72 memset(e, 0, sizeof(e)); 73 for(int j = 1; j <= n; ++j) 74 for(int k = 1; k <= n; ++k) 75 e[j] += P[i].a[k][j] * d[k]; 76 memcpy(d, e, sizeof(d)); 77 } 78 } 79 double ans = 0; 80 for (int i = 1; i <= n; ++i) { 81 v.clear(); 82 for (int j = 1; j <= n; ++j) { 83 if (j == i) continue; 84 double theta = atan2(x[j] - x[i], y[j] - y[i]); 85 while (theta < 0) theta += pi; 86 while (theta >= pi) theta -= pi; 87 v.push_back(pdi(theta, j)); 88 } 89 sort(v.begin(), v.end()); 90 for (int j = 0, k; j < n - 1;) { 91 double tmp = 0; 92 for (k = j; k < n - 1 && fabs(v[k].first - v[j].first) < eps; ++k); 93 for (int p = j; p < k; ++p) tmp += d[v[p].second] / (k - j + 1); 94 tmp += d[i] / (k - j + 1); 95 ans = max(ans, tmp); 96 j = k; 97 } 98 } 99 printf("%.10f ", ans); 100 } 101 return 0; 102 }