ABC278B枚举判断时间

#include <stdio.h>
int n, m, i, j, k, a, b, x, y;
int bad(int a, int b){
    a += b/60, a %= 24, b %= 60;
    x = a-a%10 + b/10;
    y = a%10*10 + b%10;//交换后为x:y
    if(x>23 || y>59) return 0;
    printf("%d %d\n", a, b);
    return 1;//合法则冲突
}
int main(){
    scanf("%d%d", &n, &m);
    for(i=0; i<60*24; i++){
        if(bad(n, m+i)) break;
    }//从接下来的24小时查找
    return 0;
}