Codeforces 1183H DP 计算子序列数目

题意及思路：https://blog.csdn.net/mmk27_word/article/details/93999633

第一次见这种DP，有点像退背包的思想，如果发现有可能因为字母相同和前面算重时，把这种情况减去。

代码：

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 110;
int pre[30];
LL dp[maxn][maxn];
char s[maxn];
int main() {
	int n;
	LL k, cost = 0;
	scanf("%d%lld", &n, &k);
	scanf("%s",s + 1);
	dp[0][0] = 1;
	memset(pre, -1, sizeof(pre));
	for (int i = 1; i <= n; i++) {
		int now = s[i] - ‘a‘;
		int tmp = i - pre[now];
		dp[i][0] = 1;
		for (int j = 1; j <= i; j++) {
			dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
			if(pre[now] != -1 && j >= tmp) dp[i][j] -= dp[pre[now] - 1][j - tmp];
			dp[i][j] = min(dp[i][j], k);
		}
		pre[now] = i;
	}
	for (int i = 0; i <= n; i++) {
		LL tmp = min(dp[n][i], k);
		cost += tmp * i;
		k -= tmp;
		if(k == 0) break;
	}
	if(k > 0) printf("-1\n");
	else printf("%lld\n", cost);
}

原文地址：https://www.cnblogs.com/pkgunboat/p/11105299.html