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_queue typedef 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; }