解题思路:
F[i][j] 为用了前i种花,从左到右摆了j盆的方法数。
F[i][j] = F[i-1][j] + F[i][j-1] - F[i-1][j-ai-1]
F[i-1][j-ai-1] 转移到 F[i][j]需要摆ai+1盆花,所以需要减去
注意事项:
参考代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
using namespace std;
#define mod 1000007
const int maxn = 105;
int n, m;
int a[maxn];
int f[maxn][maxn];
int main() {
cin >> n >> m;
for (int i=1;i<=n;++i) cin >> a[i];
memset(f, 0, sizeof(f));
for (int i=0;i<=n;++i) f[i][0] = 1;
for (int i=1;i<=n;++i)
for (int j=1;j<=m;++j) {
f[i][j] = f[i-1][j]+f[i][j-1];
if (j - a[i] - 1 >= 0)
f[i][j] -= f[i-1][j-a[i]-1];
f[i][j] = (f[i][j] + mod)%mod;
}
cout << f[n][m];
return 0;
}
0.0分
0 人评分