HDOJ 题目3076 ssworld VS DDD（概率dp）

发布时间：2020-12-13 22:35:28 所属栏目：百科来源：网络整理

导读：ssworld VS DDD Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1607Accepted Submission(s): 330 Problem Description One day,sssworld and DDD play games together,but there are some special

ssworld VS DDD

Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1607Accepted Submission(s): 330

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

Source

2009 Multi-University Training Contest 17 - Host by NUDT

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题目大意：两个人掷筛子，每个人都有血量h1,h2,也有掷6个数的概率，掷一次，点小的血量减一，最后谁的血量先变为0，就输了，求第一个人赢的概率，

ps：这个题的数据有坑，，，，开始一直wa，看了讨论组才知道，，数据反了，，正常应该是输入 h1,h2的，，换了一下顺序就过了。。坑啊

ac代码

#include<stdio.h>
#include<string.h>
double a[10],b[10],dp[2005][2005];
int main()
{
	int h1,h2;
	while(scanf("%d%d",&h2,&h1)!=EOF)
	{
		int i,j;
		double p1,p2,p,w1,w2;
		for(i=1;i<=6;i++)
			scanf("%lf",&a[i]);
		for(i=1;i<=6;i++)
			scanf("%lf",&b[i]);
		p1=p2=0;
		for(i=2;i<=6;i++)
		{
			for(j=1;j<i;j++)
			{
				p1+=a[i]*b[j];
				p2+=a[j]*b[i];
			}
		}
		p=1-p2-p1;
		if(p==1)
			w1=w2=0;
		else
		{
			w1=p1/(1-p);
			w2=p2/(1-p);
		}
		memset(dp,sizeof(dp));
		dp[0][0]=1;
		for(i=0;i<h2;i++)
		{
			for(j=0;j<=h1;j++)
			{
				if(i)
					dp[i][j]+=dp[i-1][j]*w1;
				if(j)
					dp[i][j]+=dp[i][j-1]*w2;
			}
		}
		double ans=0;
		for(i=0;i<h1;i++)
		{
			ans+=dp[h2-1][i]*w1;
		}
		printf("%.6lfn",ans);
	}
}

（编辑：李大同）

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