HDU 4906 Our happy ending (状压DP)

HDU 4906 Our happy ending

题目链接

题意：给定n个数字，每一个数字能够是0-l,要选当中一些数字。然后使得和为k，问方案

思路：状压dp。滚动数组，状态表示第i个数字。能组成的数字状态为s的状态，然后每次一个数字，循环枚举它要选取1 - min(l,k)的多少，然后进行状态转移

代码：

#include <cstdio>
#include <cstring>

typedef long long ll;

const int N = (1<<20) + 5;
const ll MOD = 1000000007;
int t, n, k;
ll l, dp[N];

int main() {
    scanf("%d", &t);
    while (t--) {
	scanf("%d%d%lld", &n, &k, &l);
	int s = (1<<k);
	if (l > k) {
	    ll yu = l - k;
	    l = k;
	}
	memset(dp, 0, sizeof(dp));
	dp[0] = 1;
	while (n--) {
	    for (int i = s - 1; i >= 0; i--) {
		if (dp[i] == 0) continue;
		ll tmp = yu * dp[i] % MOD;
		ll now = dp[i];
		for (int j = 1; j <= l; j++) {
		    int next = i|((i<<j)&(s - 1)|(1<<(j - 1)));
		    dp[next] = (dp[next] + now) % MOD;
		}
		dp[i] = (dp[i] + tmp) % MOD;
	    }
	}
	ll ans = 0;
	for (int i = 0; i < s; i++) {
	    if (i&(1<<(k - 1))) {
		ans = (ans + dp[i]) % MOD;
	    }
	}
	printf("%lld
", ans);
    }
    return 0;
}