Codeforces Round #427 (Div. 2)

A

懂题意就没什么问题了、

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 10;int main(){LL n, t1, t2, v1, v2;scanf("%lld%lld%lld%lld%lld", &n, &v1, &v2, &t1, &t2);LL a = t1 + v1 * n + t1;LL b = t2 + v2 * n + t2;if(a == b) {puts("Friendship");} else if(a < b) {puts("First");} else {puts("Second");}return 0;}

B

题意：原本有一个数x，它的数位之和最小是k，然后给出一个数n给你，n和x长度一样，问n和x最少有多少个对应位置的数不相同、

思路：可以这么想，如何n这个数的数位之和sum大于等于k，很显然他们两个可以相等也就是说n == x，此时它们的贡献是0，如果sum < k，此时需要贪心一下

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 10;char st[qq];int main(){int k;scanf("%d", &k);scanf("%s", st);int len = strlen(st);int sum = 0;for(int i = 0; i < len; ++i) {sum += st[i] - '0';}if(k <= sum) {puts("0");return 0;}sort(st, st + len);int cnt = sum;for(int i = 0; i < len; ++i) {cnt = cnt + (9 - (st[i] - '0'));if(cnt >= k) {printf("%d\n", i + 1);return 0;}}return 0;}

C

题意：n个点，给出坐标和初始亮度，最开始是0时刻，现在问你t时刻矩阵x1 y1 x2 y2 中亮度和是多少，每过一秒每个点的亮度加1，如果超过c则变成0

思路：dp[i][j][k]代表矩阵0 0 i j 亮度为k的点有多少个，计算一下即可

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 10;int n, q, c;LL dp[105][105][12];LL num[105][105][12];int main(){scanf("%d%d%d", &n, &q, &c);int maxnX = 100, maxnY = 100;for(int i = 0; i < n; ++i) {int x, y, z;scanf("%d%d%d", &x, &y, &z);num[x][y][z]++;dp[x][y][z]++;}for(int i = 1; i <= maxnX; ++i) {for(int j = 1; j <= maxnY; ++j) {for(int k = 0; k <= c; ++k) {dp[i][j][k] += dp[i - 1][j][k];}}}for(int i = 1; i <= maxnX; ++i) {for(int j = 1; j <= maxnY; ++j) {//printf("%d %d  QQQ ", i, j);for(int k = 0; k <= c; ++k) {dp[i][j][k] += dp[i][j - 1][k];//printf("%d ", dp[i][j][k]);}//puts("");}}while(q--) {int t, x1, x2, y1, y2;scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);t = t % (c + 1);LL sum = 0;for(int k = 0; k <= c; ++k) {LL see = (k + t) % (c + 1);//printf("%d %d %d %d\n", dp[x2][y2][k], dp[x2][y1 - 1][k], dp[x1 - 1][y2][k], dp[x1 - 1][y1 - 1][k]);LL tmp = (dp[x2][y2][k] - dp[x2][y1 - 1][k] - dp[x1 - 1][y2][k] + dp[x1 - 1][y1 - 1][k]);sum += (see * tmp);}printf("%lld\n", sum);}return 0;}

D

题意：求出回文度为1 ，2，3，...， len的个数

思路：参考：传送门

dp[i][j]代表区间[i， j]的回文度，用类似于区间dp的方法更新结果

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 5000 + 10;int dp[qq][qq];int num[qq];char st[qq];int main(){scanf("%s", st + 1);int len = strlen(st + 1);for(int l = 1; l <= len; ++l) {for(int i = 1; i + l - 1 <= len; ++i) {int j = i + l - 1;if(l == 1) {dp[i][j] = 1;continue;} else if(l == 2) {if(st[i] == st[j])dp[i][j] = 2;elsedp[i][j] = 0;continue;}if(st[i] != st[j] || !dp[i + 1][j - 1])continue;dp[i][j] = 1;int mid = l / 2;if(dp[i][i + mid - 1] && dp[j - mid + 1][j]) {dp[i][j] = dp[i][i + mid - 1] + 1;}}}for(int i = 1; i <= len; ++i) {for(int j = i; j <= len; ++j) {num[dp[i][j]]++;}}for(int i = len - 1; i >= 1; --i) {num[i] += num[i + 1];}for(int i = 1; i <= len; ++i) {printf("%d ", num[i]);}puts("");return 0;}