HDU 5194 DZY Loves Balls （组合数学）

DZY Loves Balls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 851    Accepted Submission(s): 462


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 S.

 

Input

The input consists several test cases. (TestCase≤150)

The first line contains two integers, n, 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

 

Source

BestCoder Round #35

 

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题意：袋子里有N个黑球、M个白球。用1表示取出来的是黑球，0表示取出来的是白球。不放回

的从袋子里取出这N+M个球。求相邻取出的两个球第一个球白球，第二个球是黑球的期望

次数是多少，即出现"01"的期望次数是多少。

思路：第i个位置放白球的概率m/(m+n),第i+1个位置放黑球的概率n/(m+n-1),白球能出现的位置是m+n-1，所以01的期望次数是m/(m+n)*n/(m+n-1)*(m+n-1),即（m*n)/(m+n)最后要将分子分母约分输出。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int GCD(int a,int b)
{
int t;
if(a<b)
{
t=a;
a=b;
b=t;
}
if(b==0)
return a;
return GCD(b,a%b);
}
using namespace std;
int main()
{
int a,b;
while(scanf("%d%d",&a,&b)!=EOF)
{
int m=a*b;
int n=a+b;
printf("%d/%d\n",m/GCD(m,n),n/GCD(m,n));
}
return 0;
}