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;
				else	dp[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;
}