[期望DP] BZOJ4008 [HNOI2015]亚瑟王

神犇题解：http://blog.csdn.net/popoqqq/article/details/45365759

f[i][j]应为考虑到 i 还剩 j 个机会 这样一个局面的概率

神DP

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int n,m;
long double ans;
long double p[305],d[305],f[305][305];

long double Pow(long double x,int p)
{
    long double ret=1.0;
   	for (;p;x*=x,p>>=1)
        if(p&1)
			ret*=x;
    return ret;
}

int main()
{
    freopen("t.in","r",stdin);
    freopen("t.out","w",stdout);
    int Q;
    double a,b;
    scanf("%d",&Q);
    while (Q--)
    {
        memset(f,0,sizeof f);
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&a,&b);
            p[i]=a,d[i]=b;
        }
        f[0][m]=1.0,ans=0.0;
        for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
        	{
           		f[i][j]=f[i-1][j]*Pow(1-p[i-1],j)+f[i-1][j+1]*(1-Pow(1-p[i-1],j+1));
           		ans+=f[i][j]*(1-Pow(1-p[i],j))*d[i];
        	}
        printf("%.10lf\n",(double)ans);
    }
    return 0;
}