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)
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#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;} |
Codeforces Round #196 (Div. 2) B. Routine Problem,布布扣,bubuko.com
Codeforces Round #196 (Div. 2) B. Routine Problem
原文：http://www.cnblogs.com/xiongqiangcs/p/3784616.html