Codeforces 498B Name That Tune 概率dp (看题解）

Name That Tune

刚开始我用前缀积优化dp， 精度炸炸的。

我们可以用f[ i ][ j ] 来推出f[ i ][ j + 1 ]， 记得加加减减仔细一些。。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N =  + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);

int n, T, t[N];
double dp[N][N], p[N], q[N], f[N];

int main() {
    scanf("%d%d", &n, &T);
    for(int i = ; i <= n; i++) {
        scanf("%lf%d", &p[i], &t[i]);
        p[i] /= ; q[i] =  - p[i];
        f[i] = pow(q[i], t[i]);
    }
    double ans = ;
    dp[][] = ;
    for(int i = ; i <= n; i++) {
        double gg = ;
        for(int j = ; j <= T; j++) {
            if(j < t[i]) gg = gg * q[i] + p[i] * dp[i - ][j - ];
            else if(j == t[i]) gg = gg * q[i] + p[i] * dp[i - ][j - ] + f[i] * dp[i - ][j - t[i]];
            else gg = gg * q[i] + p[i] * dp[i - ][j - ] - f[i] * dp[i - ][j - t[i] - ] + f[i] * dp[i - ][j - t[i]];
            dp[i][j] = gg;
            ans += dp[i][j];
        }
    }
    printf("%.12f\n", ans);
    return ;
}

/*
*/