Codeforces Round #302 (Div. 2) -- （A，B，C）

题目传送：Codeforces Round #302 (Div. 2)

A. Set of Strings

思路：注意开头字母都不相同

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

char s[105];

int flag[26];

int f[26];

int main() {
	int k;
	scanf("%d", &k);
	scanf("%s", s);
	int len = strlen(s);

	memset(flag, 0, sizeof(flag));
	for(int i = 0; i < len; i ++) {
		flag[s[i] - 'a'] = 1;
	}
	int cnt = 0;
	for(int i = 0; i < 26; i ++) {
		cnt += flag[i];
	}
	if(cnt < k) {
		cout << "NO" << endl;
	}
	else {
		memset(f, 0, sizeof(f));

		char cur = s[0];
		cout << "YES" << endl;
		printf("%c", s[0]);
		f[s[0] - 'a'] = 1;
		int tt = 1;
		for(int i = 1; i < len; i ++) {
			if(s[i] != cur && tt != k && f[s[i] - 'a'] == 0) {
				printf("\n%c", s[i]);
				f[s[i] - 'a'] = 1;
				tt ++;
				cur = s[i];
			}
			else printf("%c", s[i]);
		}
	}
	return 0;
}

B. Sea and Islands

思路：构造，给你一个n*n的地图，看地图上能不能有k个岛，岛是由一个联通块组成的（相邻的即为一个联通块）

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int main() {
	int n, k;
	cin >> n >> k;

	int sum = n & 1 ? (n * n + 1) / 2 : n * n / 2;//最多可以有这么多个岛 

	if(k > sum) {
		cout << "NO" << endl;
	}
	else {
		cout << "YES" << endl;
		int t = 0;
		for(int i = 1; i <= n; i++, printf("\n")) {
			for(int j = 1; j <= n; j ++) {
				if(i & 1 && t != k) {
					if(j & 1) {
						printf("L");
						t ++;
					}
					else {
						printf("S");
					}
				}
				else if(!(i & 1) && t != k){
					if(j & 1) {
						printf("S");
					}
					else {
						printf("L");
						t ++;
					}
				}
				else if(t == k) {
					printf("S");
				}
			}
		}
	}
	return 0;
}

C. Writing Code

思路：简单DP，完全背包

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <deque>
#include <cctype>
#define LL long long
#define INF 0x7fffffff
using namespace std;

LL dp[505][505];
LL a[505];

int main() {
	int n, m, b, mod;
	scanf("%d %d %d %d", &n, &m, &b, &mod);
	for(int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]);
	}
	dp[0][0] = 1;
	for(int i = 1; i <= n; i ++) {
		for(int j = 1; j <= m; j ++) {
			for(int k = 0; k <= b; k ++) {
				if(k >= a[i]) dp[j][k] += dp[j-1][k - a[i]];
				dp[j][k] %= mod;
			}
		}
	}
	LL ans = 0;
	for(int i = 0; i <= b; i ++) {
		ans += dp[m][i];
		ans %= mod;
	}
	cout << ans << endl;
	return 0;
}