Rikka with Chess
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 84 Accepted Submission(s): 72
Problem Description
Yuta gives Rikka a chess board of size n×m.
As we all know, on a chess board, every cell is either black or white and every two cells that share a side have different colors.
Rikka can choose any rectangle formed by board squares and perform an inversion, every white cell becomes black, and vice versa.
Rikka wants to turn all cells into the same color, please tell Rikka the minimal number of inversions she need to achieve her goal.
As we all know, on a chess board, every cell is either black or white and every two cells that share a side have different colors.
Rikka can choose any rectangle formed by board squares and perform an inversion, every white cell becomes black, and vice versa.
Rikka wants to turn all cells into the same color, please tell Rikka the minimal number of inversions she need to achieve her goal.
Input
The first line contains a number T(T≤10) ——The number of the testcases.
Each testcase contains two numbers n,m(n≤109,m≤109).
Each testcase contains two numbers n,m(n≤109,m≤109).
Output
For each testcase, print a single number which represents the answer.
Sample Input
3
1 2
2 2
3 3
1 2
2 2
3 3
Sample Output
1
2
2
2
2
题解：很好找的规律，第二组数据：
101
010
101
把第二行反转
101
101
101
把第二列反转
111
111
111
由此可得：
直接行列减半取整即可；
直接行列减半取整即可；
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x,y) scanf("%lf%lf",&x,&y)
#define P_ printf(" ")
const int MAXN=;
typedef long long LL;
int main(){
int T,N,M;
SI(T);
while(T--){
SI(N);
SI(M);
printf("%d\n",N/+M/);
}
return ;
}