Vasya and Triangle

ps：假设三角形的两条直角边为a，b，则 n * m / k = a * b / 2，如果等式成立，就 （2 * n * m）% k == 0，怎么构造a，b？暴力貌似不行，考虑一下gcd。注意一点：2应该乘在哪个数身上！

P p[3];
LL n, m, k;

LL gcd(LL a, LL b) {
    return (!b ? a : gcd(b, a % b));
}

int main()
{
    cin >> n >> m >> k;

    if (2 * n * m % k != 0) {
        puts("NO");
        return 0;
    }

    puts("YES");
    bool flag = (n > m);

    int g = gcd(n, k);
    n /= g, k /= g;

    int t = gcd(m, k);
    m /= t, k /= t;

    if (g != 1 && k == 1) n = 2 * n;
    else if (t != 1 && k == 1) m = 2 * m;

    p[0].first = p[0].second = 0;
    if (flag) {
        p[1].first = max(n, m), p[1].second = 0;
        p[2].first = 0, p[2].second = min(n, m);

    }
    else {
        p[1].first = min(n, m), p[1].second = 0;
        p[2].first = 0, p[2].second = max(n, m);
    }

    rep(i, 0, 3) cout << p[i].first << " " << p[i].second << endl;
    return 0;
}