HDU 5194 DZY Loves Balls（排列组合瞎搞 ）

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 isp/q(p and q are coprime)

 

Sample Input

1 12 3

 

Sample Output

1/26/5
Hint
Case 1: S='01' or S='10', so the expected times = 1/2 = 1/2Case 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
 

 

思路:先把1全部列在一条线上，枚举有一个01  两个01，当有一个01时，在n个1前面选择一个，那么这个前面和最后可以放0，也就是把0划分成两部分，同时也要保证

     前一部分有0，然后枚举01的个数就可以了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8#define fre(i,a,b)  for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define ssf(n)      scanf("%s", n)#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define bug         pf("Hi\n")using namespace std;typedef __int64 ll;#define INF 0x3f3f3f3f#define N 10005ll C(ll a,ll b){    ll x=1,y=1;    ll i;    b=min(b,a-b);    if(b==0) return 1;    fre(i,1,b+1)     {      x=x*a;      y=y*i;      a--;     }   return x/y;}ll fdd(ll x,ll y){if(y==0) return x;return fdd(y,x%y);}int main(){ll i,j;ll n,m;while(~scanf("%I64d%I64d",&n,&m)){j=min(n,m);   ll ans=0;fre(i,1,j+1)  ans=ans+i*C(n,i)*C(m,i);ll y=C(n+m,n);ll x=fdd(ans,y);pf("%I64d/%I64d\n",ans/x,y/x);}return 0;}