C语言程序设计教程（第三版）课后习题6.7 （C语言代码）简单版

超时的 把for(k=1;k<i;k++) 改成for(k=1;k<=(i/2);k++)就可以解决了

#include <stdio.h>
int main()
{
    long double N, sum = 0;
    scanf( "%d", &N );
    for ( int i = 2; i <= N; i++ )
    {
        for ( int j = 1; j <i/2; j++ )
        {
            if ( i % j == 0 )
            {
                sum = sum + j;
            }
        }
        if ( sum == i )
        {
            printf( "%d its factors are ", i );
            for ( int k = 1; k < i; k++ )
            {
                if ( i % k == 0 )printf( "%d ", k );
            }
            printf( "\n" );
        }
        sum = 0;
    }
    return(0);
}
问下我这个哪里错了？

printf("\n");
            }
           
            sum=0;                  
    }
请问这里是什么意思