Codeforces Round #196 (Div. 2) B. Routine Problem

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题目大意：给你显示器屏幕的比例(a:b)和你要看电影的比例(c:d)，求调整后剩下的没有到的区域和总面积的比例。

题目分析：在做虚拟赛的时候，对自己的想法不是很确定，能否AC。在提交WA后，就果断放弃了。第二天在做的时候发现是自己特例没想到。虽说我的想法很笨拙，但是还是能AC的。反正就是定长比宽，或者是定宽比长。

下面是代码：

#include <iostream>#include <stdio.h>using namespace std;int gcd(int a,int b){    return b == 0 ? a : gcd(b,a%b);}int main(){    int a,b,c,d;    cin>>a>>b>>c>>d;    int g1 ,g2;    g1 = gcd(a,b);    g2 = gcd(c,d);    if(g1 != 1) a /= g1,b /= g1;    if(g2 != 1) c /= g2,d /= g2;    if(a == b) a = 1,b = 1;    if(c == d) c = 1,d = 1;    double t1 = (double)a/b;    double t2 = (double)c/d;    //printf("%lf\n",t1);   // printf("%lf\n",t2);    int com;    if(t1 < t2)    {        g1 = gcd(a,c);        com = (a * c)/g1/a;        b = b*com;        com = (a*c)/g1/c;        d = d*com;        g1 = gcd(b,d);        b /= g1;        d /= g1;        printf("%d/%d\n",(b-d),b);    }    else if(t1 > t2)    {        g2 = gcd(b,d);        com = (b*d)/g2/b;        a = a * com;        com = (b*d)/g2/d;        c = c * com;        g2 = gcd(a,c);        a /= g2;        c /= g2;        printf("%d/%d\n",(a-c),a);    }    else printf("0/1\n");    return 0;}

PS：代码写的很戳....