codeforces 167B . Wizards and Huge Prize 概率dp

题目链接

dp[i][j][k]表示到第i个人赢了j个人剩余背包容量为k的情况。

然后转移就可以了。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
double dp[205][205][405], p[205];
int a[205];
int main()
{
    int n, l, tmp;
    cin>>n>>l>>tmp;
    for(int i = 1; i<=n; i++) {
        cin>>p[i];
        p[i]/=100;
    }
    for(int i = 1; i<=n; i++) {
        cin>>a[i];
    }
    dp[0][0][200+tmp] = 1;
    for(int i = 0; i<n; i++) {
        for(int j = 0; j<=i; j++) {
            for(int k = 0; k<=400; k++) {
                if(a[i+1] == -1) {
                    if(k>0)
                        dp[i+1][j+1][k-1] += dp[i][j][k]*p[i+1];
                    dp[i+1][j][k] += dp[i][j][k]*(1-p[i+1]);
                } else {
                    int val = min(400, k+a[i+1]);
                    if(k>0)
                        dp[i+1][j+1][val] += dp[i][j][k]*p[i+1];
                    dp[i+1][j][k] += dp[i][j][k]*(1-p[i+1]);
                }
            }
        }
    }
    double ans = 0;
    for(int i = l; i<=n; i++) {
        for(int j = 200; j<=400; j++) {
            ans += dp[n][i][j];
        }
    }
    printf("%.8f
", ans);
    return 0;
}