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

#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 = 5000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

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 = 1; i <= n; i++) {
        scanf("%lf%d", &p[i], &t[i]);
        p[i] /= 100; q[i] = 1 - p[i];
        f[i] = pow(q[i], t[i]);
    }
    double ans = 0;
    dp[0][0] = 1;
    for(int i = 1; i <= n; i++) {
        double gg = 0;
        for(int j = 1; j <= T; j++) {
            if(j < t[i]) gg = gg * q[i] + p[i] * dp[i - 1][j - 1];
            else if(j == t[i]) gg = gg * q[i] + p[i] * dp[i - 1][j - 1] + f[i] * dp[i - 1][j - t[i]];
            else gg = gg * q[i] + p[i] * dp[i - 1][j - 1] - f[i] * dp[i - 1][j - t[i] - 1] + f[i] * dp[i - 1][j - t[i]];
            dp[i][j] = gg;
            ans += dp[i][j];
        }
    }
    printf("%.12f\n", ans);
    return 0;
}

/*
*/