【bzoj4008】[HNOI2015]亚瑟王 期望dp

对于每个物品，把r轮放在一起考虑
f[i][j]表示到第i个物品还剩j轮的概率
f[i][j]=f[i-1][j]*pow[i-1][j]+f[i-1][j+1]*(1-pow[i-1][j+1])
ans=∑∑f[i][j]*(1-pow[i][j])*d[i]
pow[i][j]=(1-p[i])^j

还要理解一下，dp好难呀！！！

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>

using namespace std;

long double f[250][150],p[250][150],d[250];
int T,n,m,r;

int main()
{
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d%d",&n,&r);
		for (int i=1;i<=n;i++)
		{
			double x,y;
			scanf("%lf%lf",&x,&y);
			p[i][1]=1-x;d[i]=y;
		}
		p[0][1]=1;
		for (int i=0;i<=n;i++)
		  for (int j=2;j<=r+1;j++)
		    p[i][j]=p[i][j-1]*p[i][1];
		memset(f,0,sizeof(f));
		f[0][r]=1;
		for (int i=1;i<=n;i++)
		  for (int j=1;j<=r;j++)
		    f[i][j]=f[i-1][j]*p[i-1][j]+f[i-1][j+1]*(1-p[i-1][j+1]);
		long double ans=0;
		for (int i=1;i<=n;i++)
		  for (int j=1;j<=r;j++)
		    ans+=f[i][j]*(1-p[i][j])*d[i];
		printf("%.10lf\n",(double)ans);
	}
	return 0;
}