BZOJ 4008 亚瑟王(概率DP 奥妙重重)

题意

中文题面，就不解释了

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分析

显然这道题直接求期望太麻烦，想想转化问题（这转化太神了）。

定义f(i,j)f(i,j)f(i,j)表示第iii张卡总共被经过jjj次的概率，有转移方程式

f(i,j)=f(i−1,j)∗(1−pi−1)j+f(i−1,j+1)∗(1−(1−pi−1)j+1)\large f(i,j)=f(i-1,j)*(1-p_{i-1})^j+f(i-1,j+1)*(1-(1-p_{i-1})^{j+1})f(i,j)=f(i−1,j)∗(1−pi−1​)j+f(i−1,j+1)∗(1−(1−pi−1​)j+1)最终答案就是

∑i=1n∑j=1rf(i,j)∗(1−(1−p[i])j)∗di\large \sum_{i=1}^n\sum_{j=1}^rf(i,j)*(1-(1-p[i])^j)*d_ii=1∑n​j=1∑r​f(i,j)∗(1−(1−p[i])j)∗di​

−.−-.-−.−

AC CODE

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 250;
const int MAXM = 150;
double p[MAXN], d[MAXN], f[MAXN][MAXM], mul[MAXN][MAXM];
int main () {
	int T, n, m;
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d", &n, &m);
		for(int i = 1; i <= n; ++i) {
			scanf("%lf%lf", &p[i], &d[i]);
			mul[i][0] = 1;
			for(int j = 1; j <= m; ++j)
				mul[i][j] = mul[i][j-1] * (1-p[i]);
		}
		memset(f, 0, sizeof f);
		f[1][m] = 1; double ans = 0;
		for(int i = 1; i <= n; ++i)
			for(int j = m; j >= 1; --j) {
				if(!(i == 1 && j == m))
					f[i][j] = f[i-1][j] * mul[i-1][j] + f[i-1][j+1] * (1-mul[i-1][j+1]);
				ans += f[i][j] * (1-mul[i][j]) * d[i];
			}
		printf("%.10lf\n", ans);
	}
}

TIP:预处理幂快的多