Codeforces 1058 D. Vasya and Triangle 分解因子

传送门：http://codeforces.com/contest/1058/problem/D

题意：

　　在一个n*m的格点中，问能否找到三个点，使得这三个点围成的三角形面积是矩形的1/k。

思路：

　　这个题就是找（0，0）（a，0）（0，b）中的a和b，可以得到2*n*m/k = a*b。所以2*n*m%k != 0答案就不存在。接下来就是把等式左边的分母消去，得到a，b的分配值。

如果k是偶数，那个k肯定可以和2约分，所以把k除2. 再得到g = gcd（n，k），a = n/g，就是说能用n约掉一部分k就约掉，再用k = k/g，b = m/k。

如果k是奇数，等式左边的2不能约掉，就要在经过和上面相同的操作后，把a * 2或者把b*2，肯定是有一个满足不超过限制的，因为之前a或b一定除了一个大于2的数。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queuetypedef long long ll;
typedef unsigned long long ull;typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c); 
//priority_queue<int ,vector<int>, greater<int> >que;const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
// const int mod = 998244353;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;template<typename T>
inline T read(T&x){x=0;int f=0;char ch=getchar();while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x=f?-x:x;
}/*-----------------------showtime----------------------*/ll gcd(ll a,ll b){if(b == 0)return a;return gcd(b,a%b);}  
int main(){ll n,m,k;cin>>n>>m>>k;if(2*n*m%k!=0){puts("NO");return 0;}ll a,b;if(k % 2 == 0){k/=2;ll g = gcd(n,k);a = n/g;k = k / g;b = m / k;}else {ll g = gcd(n,k);a = n / g;k = k / g;b = m / k;if(2ll*a < n)a *=2ll;else b *= 2ll;}puts("YES");cout<<"0 0"<<endl;cout<<a<<" 0"<<endl;cout<<"0 "<<b<<endl;return 0;
}