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

screen 尺寸为a:b

video 尺寸为 c:d

如果a == c 则 面积比为 cd/ab=ad/cb （ad < cb）

如果b == d 则 面积比为 cd/ab=cb/ad  (cb < ad)

如果不相等时

如果a/b > c/d，则ad/bd > cb/db 则(ad > cb)

　　screen尺寸可为 ad：bd， video的尺寸可为 cb:db

　　面积比为：cb*db/ad*bd = cb/ad (ad > cb)

如果a/b < c/d，则ac/bc < ca/da 则(ad < cb)

　　screen尺寸可为 ac：bc， video的尺寸可为 ca:da

　　面积比为：ca*da/ac*bc = ad/cb (ad < cb)

 综合得面积比为min(ad,cb)/max(ad,cb)

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int gcd(int n, int m){
	while(m){
		int t = n%m;
		n = m;
		m = t;
	}
	return n;
}

int main(){
	int a,b,c,d;
	cin >> a >> b >> c >> d;
	int video = c*b, screen = a*d;
	if(video > screen) swap(screen,video);
	int k = gcd(screen,video);
	video/=k;screen/=k;
	cout<<screen-video<<"/"<<screen<<endl;
}