Codeforces Round #239 (Div. 2), problem: (C) Triangle solution

Codeforces Round #239 (Div. 2), problem: (C) Triangle: http://codeforces.com/contest/408/problem/C

Codeforces Round #239 (Div. 2), problem: (C) Triangle editorial: http://codeforces.com/blog/entry/11333

Codeforces Round #239 (Div. 2), problem: (C) Triangle solution: http://ideone.com/KVM6Cb

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

int gcd(int u, int v) {
    if(u==v) return u;
    if(u==0) return v;
    if(v==0) return u;
    if(~u & 1) {
        if(v & 1) return gcd(u>>1, v);
        else return gcd(u>>1, v>>1)<<1;
    }
    if(~v & 1) return gcd(u, v>>1);
    if(u>v) return gcd((u-v)>>1, v);
    return gcd((v-u)>>1, u);
}

int main() {
    int a, b, x, y, g, Bx, By, B;
    scanf("%d%d", &a, &b);
    for(int i=0; i<a; i++) {
        for(int j=0; j<a; j++) {
            if(pow(i, 2) + pow(j, 2)==pow(a, 2)) {
                x=i, y=j;
                g=gcd(x, y);
                Bx=-y/g;
                By=x/g;
                B=sqrt(pow(Bx, 2) + pow(By, 2));
                if(b%B==0 && sqrt(pow(By*b/B-y, 2))!=0 && sqrt(pow(Bx*b/B-x, 2))!=0) {
                    printf("YES\n");
                    printf("0 0\n");
                    printf("%d %d\n", x, y);
                    printf("%d %d\n", Bx*b/B, By*b/B);
                    return 0;
                }
            }
        }
    }
    printf("NO");
    return 0;
}

note: the above code is my solution based on the editorial. you can have a look for another sol's also. i like the below solutions also. check other's codes if you dont get editorial well.

By BigChampion, contest: Codeforces Round #239 (Div. 1), problem: (A) Triangle, Accepted, #

---

#include <cstdio>
using namespace std;
int gcd(int x,int y)
{
 if (!y) return x; return gcd(y,x%y);
}
int main()
{
 int a,b,x=0,y,i,j;
 scanf("%d%d",&a,&b);
 int g=gcd(a,b);
 for (i=1;i<g;i++)
     for (j=1;j<g;j++) if ((i*a!=j*b)&&(i*i+j*j==g*g)) x=i,y=j;
 if (!x) {printf("NO\n"); return 0;}
 printf("YES\n");
 printf("1 1\n%d %d\n%d %d\n",1+a/g*x,1+a/g*y,1+b/g*y,1-b/g*x);
 return 0;
}

By ehsan.m.a.e, contest: Codeforces Round #239 (Div. 2), problem: (C) Triangle, Accepted, #

---

#include<iostream>
#include <cmath>
using namespace std;
int main()
{
int a,b,i,a2;
float t,ta,tb;
cin>>a>>b;
a2=a*a;
for(i=1;i<a;i++)
{
    t=sqrt(a2-i*i);
    if(t-(int)t==0)
    {
        ta=-1*(b*t)/a;
        tb=(b*i)/a;
        if(ta-(int)ta==0 && tb-(int)tb==0 && tb!=t)
        {
            cout<<"YES\n0 0\n"<<i<<" "<<t<<"\n"<<ta<<" "<<tb;
            return 0;
        }
    }
}
cout<<"NO";
return 0;
}