hdu 5194 DZY Loves Balls（暴力，数学期望）

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本文介绍了一个关于从盒子中随机取出黑白球形成序列的问题，并探讨了序列中特定子序列“01”出现次数的数学期望。通过列举所有可能情况的方法给出了问题的解答，并提出了一种更高效的解决方案。

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DZY Loves Balls

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 856 Accepted Submission(s): 466

Problem Description

There are

n black balls and

m white balls in the big box.

Now, DZY starts to randomly pick out the balls one by one. It forms a sequence

S. If at the

i-th operation, DZY takes out the black ball,

Si=1, otherwise

Si=0.

DZY wants to know the expected times that '01' occurs in

Input

The input consists several test cases. (

TestCase≤150)

The first line contains two integers,

m(1≤n,m≤12)

Output

For each case, output the corresponding result, the format is

p/q(

p and

q are coprime)

Sample Input

1 1
2 3

Sample Output

1/2
6/5
HintCase 1: S='01' or S='10', so the expected times = 1/2 = 1/2
Case 2: S='00011' or S='00101' or S='00110' or S='01001' or S='01010' 
or S='01100' or S='10001' or S='10010' or S='10100' or S='11000',
so the expected times = (1+2+1+2+2+1+1+1+1+0)/10 = 12/10 = 6/5

题意:有n个1和m个0，问组成的所有串中01的数学期望是多少？

思路：最简单的思路就是暴力枚举出所有串，然后对a和b求一次gcd即可，n和m都是12，复杂度可以承受。

更加高效方便的办法是利用期望的可加性，把每一个位置上01串的期望相加即可得到总期望值

代码1：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
long long a;
int n,m;
void dfs(int s,int t,int d,int last,int num)
{
    if(d==n+m)
    {
        a+=num;
        return;
    }
    if(s<n)  dfs(s+1,t,d+1,0,num);
    if(t<m)
    {
        if(last==0) dfs(s,t+1,d+1,1,num+1);
        else dfs(s,t+1,d+1,1,num);
    }
}
long long gcd(long long a,long long b)
{
    return b?gcd(b,a%b):a;
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        a=0;
        dfs(0,0,0,-1,0);
        long long b=1;
        for(long long i=n+m; i>=m+1; i--)
            b*=i;
        for(long long i=n; i; i--)
            b/=i;
        long long c=gcd(a,b);
        printf("%lld/%lld\n",a/c,b/c);
    }
    return 0;
}

代码2：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
long long a;
int n,m;
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        int a=n*m,b=n+m;
        int c=gcd(a,b);
        printf("%d/%d\n",a/c,b/c);
    }
    return 0;
}