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\n",ans);    }    return 0;}