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：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int inf = 0x3f3f3f3f;
 5 const int maxn = 2e3 + 10;
 6 #define lowbit(x) x&-x
 7 ll gcd(ll a, ll b) {
 8     return b?gcd(b,a%b):a;
 9 }
10 int main() {
11     ll n, m, k;
12     scanf("%lld %lld %lld", &n, &m, &k);
13     if(2ll * n * m %k != 0) {
14         puts("NO");
15         return 0;
16     }
17     puts("YES");
18     puts("0 0");
19     ll x2, y2, x3, y3;
20     ll sum = 2ll * n * m / k;
21  //   printf("%lld\n", gcd(2ll*n,k));
22     if(gcd(2ll*n,k) == 1) {
23         ll u = gcd(2ll*m, k);
24         printf("%lld %lld\n%lld %lld", n/(k/u), 0ll, 0ll, 2ll*m/u);
25     } else {
26         ll u = gcd(2ll*n, k);
27         printf("%lld %lld\n%lld %lld", 2ll*n/u, 0ll, 0ll, m/(k/u));
28     }
29     return 0;
30 }