守卫挑战

金牌导航 期望-9

题目大意

有n个数，到第i个数，有p_i的概率选择这个数，问你最后选了最少L个数，且选的数的和再加k大于等于0

样例输入

3 1 0
10 20 30
-1 -1 2

样例输出

0.300000

数据范围

$0⩽k⩽2000$
$0⩽L⩽n⩽200$
$-1⩽a_i⩽1000$
$0⩽p_i⩽100$

解题思路

设$f_{i,j,k}$为前i个数，选了j个，且数的和为k的期望值
那么有：
$f_{i,j,k}=f_{i-1,j,k}\times (1-p_i)$
$f_{i,j,k}+=f_{i-1,j-1,k-a[i]}\times p_i$
直接dp即可

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define g(x) (x + 200)//使所有数大于0方便存储
#define max(x, y) ((x)>(y)?(x):(y))
using namespace std;
int n, l, K, a[210];
double ans, p[210], f[210][210][450];
int main()
{scanf("%d%d%d", &n, &l, &K);for (int i = 1; i <= n; ++i){scanf("%lf", &p[i]);p[i] /= 100.0;}for (int i = 1; i <= n; ++i)scanf("%d", &a[i]);f[0][0][g(K)] = 1.0;for (int i = 1; i <= n; ++i)for (int j = 0; j <= i; ++j){for (int k = -200; k <= 200; ++k){f[i][j][g(k)] = f[i - 1][j][g(k)] * (1.0 - p[i]);if (j > 0 && -200 <= k - a[i] && k - a[i] <= 200) f[i][j][g(k)] += f[i - 1][j - 1][g(k - a[i])] * p[i]; //dp}for (int k = max(-200, 200 - a[i]); k <= 200; ++k)f[i][j][401] += f[i - 1][j - 1][g(k)] * p[i];//高于200的不可能再变成负数，就计算到一起f[i][j][401] += f[i - 1][j][401];}for (int i = l; i <= n; ++i)for (int j = 0; j <= 201; ++j)ans += f[n][i][g(j)];//计算结果printf("%.6lf", ans);return 0;
}