bzoj 4832 抵制克苏恩 概率期望dp

考试时又翻车了.....

一定要及时调整自己的思路！！！

随从最多有7个，只有三种，所以把每一种随从多开一维

so：f[i][j][k][l]为到第i次攻击前，场上有j个1血，k个2血，l个3血随从的概率

最后利用期望的可加性都加起来就好了

ps.30滴血受到四五十伤害，完全tm不符合逻辑啊，mdzz！！！

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int T,a,b,c,n;
double f[55][8][8][8],ans;
int main()
{
    //freopen("defcthun.in","r",stdin);
    //freopen("defcthun.out","w",stdout);
    scanf("%d",&T);
    for(int i=1;i<=T;i++)
    {
        scanf("%d%d%d%d",&n,&a,&b,&c);
        ans=0; memset(f,0,sizeof f);
        f[1][a][b][c]=1;
        for(int i=1;i<=n;i++)
            for(int j=0;j<=7;j++)
                for(int k=0;k<=7;k++)
                    for(int l=0;l<=7;l++){
                        if(!f[i][j][k][l]) continue;
                        ans+=(double)f[i][j][k][l]*1.0/(1.0+j+k+l);
                        if(i==n) continue;
                        f[i+1][j][k][l]+=(double)f[i][j][k][l]*1.0/(1.0+j+k+l);
                        if(j>0) f[i+1][j-1][k][l]+=(double)f[i][j][k][l]*j/(1.0+j+k+l);
                        if(k>0){
                            if(j+k+l==7) f[i+1][j+1][k-1][l]+=(double)f[i][j][k][l]*k/(1.0+j+k+l);
                            else f[i+1][j+1][k-1][l+1]+=(double)f[i][j][k][l]*k/(1.0+j+k+l);
                        }
                        if(l>0){
                            if(j+k+l==7) f[i+1][j][k+1][l-1]+=(double)f[i][j][k][l]*l/(1.0+j+k+l);
                            else f[i+1][j][k+1][l]+=(double)f[i][j][k][l]*l/(1.0+j+k+l);
                        }
                    }
        printf("%0.2lf
",ans);
    }
    return 0;
}