TOJ1169

http://acm.tju.edu.cn/toj/showp1169.html

这道题比较简单，先输入一个int型二维数组，然后分别对其每行每列求和，结果存入两个新的数组。再判断这两个数组中偶数的个数，偶数应该是只能为偶数个（以4*4为例，偶数的可能个数有：0 2 4 6 8）。分别对偶数个数的不同情况进行输出即可。

代码：

#include 
#include 

int matrix[101][101];


int main(){
    int N;
    int horizon[101];
    int vertical[101];


    int x, y, judge_x, judge_y;
    while(scanf("%d", &N) && N){
        judge_x = 0;
        judge_y = 0;
        for(int i = 0; i < 101; i++){
            horizon[i] = 0;
            vertical[i] = 0;
        }
        for(int i = 0; i < N; i++){
            for(int j = 0; j < N; j++){
                scanf("%d", &matrix[i][j]);
                //printf("%d ", matrix[i][j]);
                //求每一横行的和
                horizon[i] += matrix[i][j];
            }
            //printf("%d ", horizon[i]);

        }
        //求每一竖行的和
        for(int j = 0; j < N; j++){
            for(int i = 0; i < N; i++){
                vertical[j] += matrix[i][j];
            }
        }

        for(int i = 0; i < N; i++){
            if(horizon[i] % 2 == 0)
                judge_x++;
            else
                x = i+1;
            if(vertical[i] % 2 == 0)
                judge_y++;
            else
                y = i+1;
        }
        if(judge_x == N && judge_y == N)    printf("OK\n");
        else if(judge_x < N-1 || judge_y < N-1)   printf("Corrupt\n");
        else if(judge_x == N-1 && judge_y == N-1) printf("Change bit (%d,%d)\n", x, y);

    }
    return 0;
}