GCJ 2008 APAC local onsites C Millionaire

时间复杂度很大。dp[i][j]表示第i轮 j这种状态的概率。

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

double dp[20][50000];
int T;
int M;
int X;
double P;

void work()
{
    memset(dp,0,sizeof dp);
    dp[0][1<<M]=1.00;
    for(int r=1; r<=M; r++)
    {
        for(int j=0; j<=(1<<M); j++)
        {
            double ans=0;
            for(int k=0;; k++)
            {
                if(j+k>(1<<M)||j-k<0) break;
                ans=max(ans,P*dp[r-1][j+k]+(1-P)*dp[r-1][j-k]);
            }
            dp[r][j]=ans;
        }
    }
}

int main()
{
    //freopen("C-large-practice.in","r",stdin);
// freopen("C-large-practice.out","w",stdout);

    scanf("%d",&T);
    for(int Case=1; Case<=T; Case++)
    {
        scanf("%d%lf%d",&M,&P,&X);
        work();
        long long XX=(long long)X;
        int d=(1<<M);
        long long dd=(long long)d;
        long long RR=XX*dd/(long long)1000000;
        int R=int(RR);
        printf("Case #%d: %6lf\n",Case,dp[M][R]);
    }
    return 0;
}

 

posted @ 2016-03-18 21:25  Fighting_Heart  阅读(511)  评论(0编辑  收藏  举报