Problem:
Score : 200 points

Problem Statement

We have a grid with N rows and M columns of squares. Initially, all the squares are white.

There is a button attached to each row and each column.
When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa.
When a button attached to a column is pressed, the colors of all the squares in that column are inverted.

Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.

Constraints

1 \leq N,M \leq 1000

0 \leq K \leq NM

Input

Input is given from Standard Input in the following format:

N M K

Output

If Takahashi can have exactly K black squares in the grid, print Yes; otherwise, print No.

Sample Input 1

2 2 2

Sample Output 1

Yes

Press the buttons in the order of the first row, the first column.

Sample Input 2

2 2 1

Sample Output 2

No

Sample Input 3

3 5 8

Sample Output 3

Yes

Press the buttons in the order of the first column, third column, second row, fifth column.

Sample Input 4

7 9 20

Sample Output 4

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp> // Common file
//#include <ext/pb_ds/tree_policy.hpp> // Including seg_tree_order_statistics_node_update
#include <stdio.h>
#include <cassert>
using namespace std;
//using namespace __gnu_pbds;
typedef long long lo;
typedef pair<lo,lo> ll;//pair
typedef vector<lo> vl;  //vector of long
typedef vector<ll > vll;   //vector of pair
typedef priority_queue<lo>p_q;
typedef vector< vl > vvl;  //vector of vectors
#define X first
#define Y second
#define mp(a,b) make_pair((a),(b))
#define REP(a,b) for(lo i=(a);i<(lo)b;i++)//no need to declare variable i
#define REPE(a,b,c,d) REP(a,b)for(lo j=(c);j<(lo)d;j++)//no need to declare vaiables i,j
#define REPV(a,b,c) for(lo(a)=b;(a)<(c);(a)++)//a is the variable
#define IREP(a,b) for(lo i=(a);i>=(b);i--)
#define IREPV(a,b,c) for(lo(a)=b;(a)>=(c);(a)--)
#define all(v) (v).begin(),(v).end()
#define TRV(a) for(auto it : a)
#define INF 100001
#define MOD lo(1e17)
#define M 1000000007
#define CHECK_BIT(var,pos) ((var) & (1<<(pos)))
#define pb(a) push_back((a))
#define endl "\n"
#define eps 1e-9
#define PI 3.141592653589793
template<typename T> ostream& operator<< ( ostream &o,vector<T> v ) {
    if ( v.size() >0 )o<<v[0];
    for ( unsigned   i=1; i<v.size(); i++ )o<<" "<<v[i];
    return o<<endl;
}
template<typename U,typename V> ostream& operator<< ( ostream &o,pair<U,V> p ) {
    return o<<"("<<p.first<<", "<<p.second<<") ";
}
template<typename T> istream& operator>> ( istream &in,vector<T> &v ) {
    for ( unsigned   i=0; i<v.size(); i++ )in>>v[i];
    return in;
}
template<typename T> istream& operator>> ( istream &in,pair <T,T> &p ) {
    in>>p.X;
    in>>p.Y;
    return in;
}
int main(){
    std::ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);cout.precision(20);
    lo n,m,k;
    cin>>n>>m>>k;
    bool ans=false;
    if(k%n==0 and k/n<=m)ans=true;
    if(k%m==0 and k/m<=n)ans=true;
    if(k==0)ans=true;
    if(ans){
        cout<<"Yes";
        return 0;
    }
    if(!ans)REP(0,m)if((k-i*n)%(m-i)==0 and (k-i*n)/(m-i)<=n and k-i*n>=0)ans=true;
    if(k<m and k<n)ans=false;
    if(k>m*n)ans=false;
    if(ans)cout<<"Yes";
    else cout<<"No";
    return 0;
}