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
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
1 #include<iostream>
2 #include<cstring>
3 #include<string>
4 #include<cstdio>
5 #include<vector>
6 #include<cmath>
7 #include<stack>
8 #include<set>
9 #include<queue>
10 #include<algorithm>
11 using namespace std;
12 #define in 1000000007
13 int main()
14 {
15 int n,m,p;
16 cin>>n>>m>>p;
17 set<int >s;
18 s.insert(0);
19 for(int i=1;i<=n;i++)
20 {
21 s.insert(i*m);
22 for(int j=1;j<=m;j++)
23 {
24 int t=i*m+j*(n-2*i);
25 s.insert(t);
26 }
27 }
28 for(int j=1;j<=m;j++)
29 {
30 s.insert(j*n);
31 for(int i=1;i<=n;i++)
32 {
33 int t=j*n+i*(m-2*j);
34 s.insert(t);
35 }
36 }
37 if(s.count(p)) cout<<"Yes"<<endl;
38 else cout<<"No"<<endl;
39 return 0;
40 }