HDOJ2503 ( a/b + c/d ) 【最大公约数GCD，最小公倍数LCM】

 

Problem : 2503 ( a/b + c/d )     Judge Status : Accepted
RunId : 5974062    Language : C++    Author : qq1203456195
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta
 1 #include <cstdio>
 2 #include <iostream>
 3 using namespace std;
 4 int gcd(int a,int b)
 5 {
 6     int t,min,max;
 7     min=a<b?a:b;
 8     max=a<b?b:a;
 9     while (t=max%min,t!=0){    max=min;min=t;    }
10     return min;
11 }
12 int lcm(int a,int b)
13 {
14     int min,max,ret;
15     min=a<b?a:b;
16     max=a<b?b:a;
17     ret=max;
18     while (ret%min!=0)    ret+=max;
19     return ret;
20 }
21 int main()
22 {
23     int cas,a,b,c,d,f,e,g;
24     scanf("%d",&cas);
25     while (cas--)
26     {
27         scanf("%d%d%d%d",&a,&b,&c,&d);
28         f=lcm(b,d);
29         e=a*(f/b)+c*(f/d);
30         g=gcd(e,f);
31         printf("%d %d\n",e/g,f/g);
32     }
33     return 0;
34 }

 

posted @ 2012-05-20 22:37  ZH奶酪  阅读(556)  评论(0)  编辑  收藏  举报