HDU 4906 Our happy ending (状压DP)
2017-05-20 20:40
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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 ; 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\n", ans); } return 0; }
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