HDU 3076 ssworld VS DDD 【概率dp】

最新推荐文章于 2019-03-09 21:14:11 发布

Puya

最新推荐文章于 2019-03-09 21:14:11 发布

阅读量341

点赞数

CC 4.0 BY-SA版权

分类专栏： HDU 动态规划

本文链接：https://blog.youkuaiyun.com/Puyar_/article/details/74066906

HDU 同时被 2 个专栏收录

175 篇文章

订阅专栏

动态规划

44 篇文章

订阅专栏

本文介绍了一个游戏中的概率计算问题，玩家通过掷骰子减少对方的生命值（HP），目标是预测一个玩家获胜的概率。文章提供了一段代码，实现了根据不同掷骰子概率来计算玩家获胜的概率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

ssworld VS DDD

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2646 Accepted Submission(s): 530

Problem Description

One day, sssworld and DDD play games together, but there are some special rules in this games.
They both have their own HP. Each round they dice respectively and get the points P1 and P2 (1 <= P1, P2 <= 6). Small number who, whose HP to reduce 1, the same points will remain unchanged. If one of them becomes 0 HP, he loses.
As a result of technical differences between the two, each person has different probability of throwing 1, 2, 3, 4, 5, 6. So we couldn’t predict who the final winner.

Input

There are multiple test cases.
For each case, the first line are two integer HP1, HP2 (1 <= HP1, HP2 <= 2000), said the first player sssworld’s HP and the second player DDD’s HP.
The next two lines each have six floating-point numbers per line. The jth number on the ith line means the the probability of the ith player gets point j. The input data ensures that the game always has an end.

Output

One float with six digits after point, indicate the probability sssworld won the game.

Sample Input

  
   5 5
1.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 1.000
5 5
0.000 0.000 0.000 0.000 0.000 1.000
1.000 0.000 0.000 0.000 0.000 0.000

Sample Output

  
   0.000000
1.000000

Source

2009 Multi-University Training Contest 17 - Host by NUDT

Recommend

lcy

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b)  memset(a,b,sizeof(a))
const int M=1e3+10;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
int i,j,k,n,m;
double a[M],b[M];
double ans;
double aw,bw;
double dp[2001][2001];

int main()
{
    while(~scanf("%d%d",&n,&m)){
        for(int i=1;i<=6;i++)scanf("%lf",&a[i]);
        for(int i=1;i<=6;i++)scanf("%lf",&b[i]);
        aw=bw=0;
        for(int i=2;i<=6;i++)
        for(int j=1;j<i;j++){
            aw+=a[i]*b[j];
            bw+=b[i]*a[j];
        }
        double p1=aw;
        double p2=bw;
        double p=1-p1-p2;
        if(p==1){
            printf("0.000000\n");
            continue;
        }
        else {
            aw=p1/(1-p);
            bw=p2/(1-p);
        }
        dp[0][0]=1;
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++){
            if(i==0&&j==0)continue;
            dp[i][j]=0;
            if(i>0)dp[i][j]+=dp[i-1][j]*aw;
            if(j>0)dp[i][j]+=dp[i][j-1]*bw;

        }
        ans=0;
        for(int i=0;i<m;i++)
            ans+=dp[n-1][i]*aw;
            if(ans>1)ans=1;
        printf("%.6lf\n",ans);
    }
    return 0;
}