2016长城信息杯中国大学生程序设计竞赛中南邀请赛 xtu 1244 Gambling

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http://www.dengwenhuo.cn/?id=199

In ICPCCamp, gambling is conducted as follow. A player will draw balls from an opaque bag containingared balls,bgreen balls andcblue balls initially. Balls in the bag will be drawn in equal probability while drawn balls won't be put back into the bag. The player continues the process until he collects allared balls, or allbgreen balls, or allcblue balls. He ends up with First Prize if he collectsared balls, or Second Prize if he collectsbgreen balls, or Third Prize otherwise.

Bobo is going to take part in the above gambling. Therefore, it is very important for him to work out the probability of winning each prize.

Input

The input contains at most40sets. For each set:

Three integersa,b,c(1≤a,b,c≤103).

Output

For each set, three irreducible fractionspA,pB,pCdenote the probability of winning First Price, Second Price, and Third Price, respectively.

Sample Input

1 1 11 2 3

Sample Output

1/3 1/3 1/37/12 4/15 3/20

只能说xtu死坑。。。好些天写的了死活不对然后long long改成__int64就过了我的天 omg wtf

设红a个绿b个黄c个

那么先取完红的前题下最后一个取分两种一种是黄一种是绿那么这两种概率相加不就是红了吗？

推一推公式默认最后一个是绿那么有(（a+(b-1)+c）! /( a!*(b-1)!*c! ) ) / ( (a+b+c)! /(a!*b!*c!))

化简得 b/(a+b+c) 然后在这之前把黄球取完有（a+c-1）!/(a!*(c-1)!) /(a+c)!/(a!*c!)

化简得c/(a+c) 之两项成立就是满足先取完红再取完黄再取完绿~~~

那么最后一个取绿=b/((a+b+c) * c/(a+c) 最后一个是黄=c/(a+b+c) * b/(a+b)

两者相加然后同理可以求出先取完绿先取完黄然后化简就成了下面的这样

#pragma comment(linker, "/STACK:102400000,102400000")#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007using namespace std;ll gcd(ll b,ll a){    if(a%b==0)        return b;    return gcd(a%b,b);}int main(){    ll t1,t2,t3,x,y,z,u1,u2,u3,d1,d2,d3;    while(scanf("%I64d %I64d %I64d",&x,&y,&z)==3)    {        ll sum=x+y+z;        u1=y*z*(sum+x);        d1=sum*(x+z)*(x+y);         u2=x*z*(sum+y);        d2=sum*(x+y)*(z+y);         u3=x*y*(sum+z);        d3=sum*(x+z)*(z+y);        t1=gcd(u1,d1);        t2=gcd(u2,d2);        t3=gcd(u3,d3);        printf("%I64d/%I64d %I64d/%I64d %I64d/%I64d\n",u1/t1,d1/t1,u2/t2,d2/t2,u3/t3,d3/t3);    }    return 0;}

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