DZY Loves Balls

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

Total Submission(s): 860    Accepted Submission(s): 467

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

Hint

Case 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

 

题解：特判了下12 12以及 11 12 和 12 11..其余的话暴力枚举就行了..

#include
      
       #include
       
        #include
        
         #include 
         
          #include 
          
           using namespace std;typedef long long LL;int n,m,p,q;int ans[30];void dfs(int step,int a,int b){    if(a>m||b>n) return; ///剪枝,不然TLE    if(a==m&&b==n){        q++;        int k = 0;        for(int i=0;i
          
         
        
       
      

 改成了标记前结点，快了许多。

#include
      
       #include
       
        #include
        
         #include 
         
          #include 
          
           using namespace std;typedef long long LL;int n,m,p,q;void dfs(int a,int b,int pre,int num){    if(a>n||b>m) return;    if(a==n&&b==m){        p+=num;        q++;        return;    }    if(b<=m) dfs(a,b+1,0,num);    if(a<=n){        if(pre==0) dfs(a+1,b,1,num+1);        else dfs(a+1,b,1,num);    }}int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        p = q = 0;        dfs(0,0,-1,0);        int d = gcd(p,q);        printf("%d/%d\n",p/d,q/d);    }    return 0;}