Opposite Task | LOJ-1001 | C

* Problem Link: https://lightoj.com/problem/opposite-task
*Solution:
 

#include <stdio.h>

int main() {
    int a,i,b;
    scanf("%d",&a);
    for(i=0;i<a;i++){
        scanf("%d",&b);
        if(b>10){
            if(b%2==0){
                printf("%d %d\n",(b/2),(b/2));
            }
            else printf("%d %d\n",((b+1)/2),(((b+1)/2)-1));
        }
        else printf("%d %d\n",(b-b),(b));
    }
    return 0;
}

Post a Comment

Post a Comment (0)

Previous Post Next Post