BZOJ4832 抵制克苏恩（概率期望+动态规划）

　　注意到A+B+C很小，容易想到设f[i][A][B][C]为第i次攻击后有A个血量为1、B个血量为2、C个血量为3的期望伤害，倒推暴力转移即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
double f[][][][];
int T,n,a,b,c;
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4832.in","r",stdin);
    freopen("bzoj4832.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    for (int i=;i<=;i++)
        for (int x=;x<=;x++)
            for (int y=;y<=-x;y++)
                for (int z=;z<=-x-y;z++)
                {
                    f[i][x][y][z]=f[i-][x][y][z]+;
                    if (x) f[i][x][y][z]+=f[i-][x-][y][z]*x;
                    if (y) f[i][x][y][z]+=f[i-][x+][y-][z+(x+y+z<)]*y;
                    if (z) f[i][x][y][z]+=f[i-][x][y+][z-+(x+y+z<)]*z;
                    f[i][x][y][z]/=(x+y+z+);
                }
    while (T--)
    {
        n=read(),a=read(),b=read(),c=read();
        printf("%.2lf\n",f[n][a][b][c]);
    }
    return ;
}