CODE FESTIVAL 2017 qual A--B-fLIP（换种想法，暴力枚举）

个人心得：开始拿着题目还是有点懵逼的，以前做过相同的，不过那是按一个位置行列全都反之，当时也是没有深究。现在在打比赛不得不

重新构思，后面一想把所有的状态都找出来，因为每次确定了已经按下的行和列后，按不同的操作所加的数都是一样的，于是就想到了用set

暴力枚举，从1-n个分别行列按钮，然后再枚举不同操作即确定行时再对列进行操作，每次操作放入set就可以了。

题目：

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≤N,M≤1000
• 0≤K≤NM

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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

Copy

2 2 2

Sample Output 1

Copy

Yes

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

---

Sample Input 2

Copy

2 2 1

Sample Output 2

Copy

No

---

Sample Input 3

Copy

3 5 8

Sample Output 3

Copy

Yes

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

---

Sample Input 4

Copy

7 9 20

Sample Output 4

Copy

No

 #include<iostream>
 #include<cstring>
 #include<string>
 #include<cstdio>
 #include<vector>
 #include<cmath>
 #include<stack>
 #include<set>
 #include<queue>
 #include<algorithm>
 using namespace std;
 #define in 1000000007
 int main()
 {
     int n,m,p;
     cin>>n>>m>>p;
     set<int >s;
     s.insert();
     for(int i=;i<=n;i++)
     {
         s.insert(i*m);
         for(int j=;j<=m;j++)
         {
             int t=i*m+j*(n-*i);
             s.insert(t);
         }
     }
     for(int j=;j<=m;j++)
     {
         s.insert(j*n);
         for(int i=;i<=n;i++)
         {
             int t=j*n+i*(m-*j);
             s.insert(t);
         }
     }
     if(s.count(p)) cout<<"Yes"<<endl;
     else cout<<"No"<<endl;
     return ;
 }