HDU 相遇周期

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HDU -
相遇周期

/*Problem :相遇周期Author  : herongweiLanguage : G++CreateTime: 2016/01/11*//*解析：两周期分别为a/b、c/d，既然是相遇周期，就是最早什么时间相遇，也就是两者的最小公倍数，转化一下，也就是求LCM（a, c）/ gcd（b, d）。直接求即可，不一定要用long long 或者 __int64，int也能过。注意：先把a/b和c/d尽可能的化简再求。*/#include <math.h>#include <queue>#include <deque>#include <vector>#include <stack>#include <stdio.h>#include <ctype.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;#define Max(a,b) a>b?a:b#define Min(a,b) a>b?b:a#define mem(a,b) memset(a,b,sizeof(a))int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};const double eps = 1e-6;const double Pi = acos(-1.0);static const int inf= 0x3f3f3f3f;static const int maxn = 1e5+10;typedef long long LL;int m, n;char ss[1005][1005];typedef  pair<int, int> P;int ans[1003];int cnt;int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}int lcm(int a,int b){    return a/gcd(a,b)*b;}int main(){   // freopen("1.txt","r",stdin);    int t;    scanf("%d",&t);    while(t--)    {         int a1,b1,a2,b2;         scanf("%d/%d %d/%d",&a1,&b1,&a2,&b2);         int g1=gcd(a1,b1);         int g2=gcd(a2,b2);         a1/=g1;b1/=g1;         a2/=g2;b2/=g2;         int c1=lcm(a1,a2);         int c2=gcd(b1,b2);         int c3;         if(c1%c2==0) printf("%d\n",c1/c2);         else {            int g3=gcd(c1,c2);            c1/=g3;            c2/=g3;            printf("%d/%d\n",c1,c2);         }    }    return 0;}
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