codeforces 1030D Vasya and Triangle【思维+gcd】

题目：戳这里

题意：选出三个点构成三角形，要求面积为n*m/k。

解题思路：因为三个点的坐标都是正整数，根据三角形面积公式(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))/2=n*m/k可知，若三角形存在，则2*n*m/k必为整数。若面积*2为整数，则把该三角形放置在x轴上即可。于是设x1=0,y1=0,x2=a,y2=0,x3=0,y3=b;求a*b=2*n*m/k。(0<=a<=n,0<=b<=m)

欲找到满足条件，可以利用最大公约数。若gcd(2*n,k)==1，则gcd(2*m,k)>=2; b=2*m/gcd(2*m,k) <=m，a=n/(k/gcd(2*m,k)<=n,反之若gcd(2*n,k)>=2同理。

附ac代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 2e3 + 10;
#define lowbit(x) x&-x
ll gcd(ll a, ll b) {
    return b?gcd(b,a%b):a;
}
int main() {
    ll n, m, k;
    scanf("%lld %lld %lld", &n, &m, &k);
    if(2ll * n * m %k != 0) {
        puts("NO");
        return 0;
    }
    puts("YES");
    puts("0 0");
    ll x2, y2, x3, y3;
    ll sum = 2ll * n * m / k;
 //   printf("%lld\n", gcd(2ll*n,k));
    if(gcd(2ll*n,k) == 1) {
        ll u = gcd(2ll*m, k);
        printf("%lld %lld\n%lld %lld", n/(k/u), 0ll, 0ll, 2ll*m/u);
    } else {
        ll u = gcd(2ll*n, k);
        printf("%lld %lld\n%lld %lld", 2ll*n/u, 0ll, 0ll, m/(k/u));
    }
    return 0;
}