完数的判断 -- C 思路简单清晰 不用指针 for循环+数组搞定 有疑惑的小伙伴看过来 ~~~

感谢大佬，完美复刻
    int num, i, N, k, Sn;
    scanf("%d", &N);
    int a[1000];
    for(num=1;num<=N;num++){//scan the numbers under the N 
        for(i=1, k=0, Sn=0;i<num;i++){//find its factors and record it
            if(num%i==0){
                a[k++]=i;
                Sn = Sn + i;//count the sum of arrys
            } 
        }
        if(Sn==num){//put out result of completed number
            printf("%d its factors are", num);
            for(i=0;i<k;i++){
                printf(" %d", a[i]); 
            }
            printf("\n");
        } 
    }

这个解法，时间超限了QAQ

这个会打印输出，剩下的自己写吧
void Check(int data)
{
    int sum = 0;
    for(int i= 1;i < data;i++)
    {
        if(data % i == 0)
        {
            sum += i;
        }
    }
    if(sum == data)
    {
       printf("%d its factors are ",data);
        for(int i= 1;i < data;i++)
        {
            if(data % i == 0)
            {
                printf("%d " ,i);
            }
        }
        printf("\n");
    }
}

j<i 能不能换成 j<(i/2)

#include<stdio.h>
 int wansu(int x);
 void display (int x);
 int main()
{
 	int x,ad;
 	scanf("%d",&x);
 	for(ad=2;ad<=x;ad++)
 	{
 		if(wansu(ad))
 		{
 			display (ad); 
		}
	}
	return 0;
}
int wansu(int x)//判断是否为完数 
{
	int id,sum;
	for(id=1,sum=0;id<=(x/2);id++)
	{
		if(x%id==0)
		{
			sum=sum+id;
		}
	}
	if(sum==x) return 1;
	else return 0;
}
void display (int x)
{
	int id;
	printf("%d its factors are ",x);
	for(id=1;id<=(x/2);id++)
	{
		if(x%id==0)
		{
			printf("%d ",id);
		}
	}
	printf("\n");
}

#include<stdio.h>
#include<math.h>
#include<string.h>
int main()
{
  int n;
  scanf("%d",&n);
  int i,j,k;
  int a[100];
  int sum=0;
  int count=0;
  for(i=2;i<=n;i++){
    for(j=1;j<i;j++){
       if(i%j==0){
       sum=sum+j;
      // printf("%d ",sum);
      a[count]=j;
      count++;
       }

    }if(sum==i){printf("%d its factors are ",i);
    for(k=0;k<count;k++){
        printf("%d ",a[k]);
    }
    printf("\n");
    }
    sum=0;
    count=0;
  }

}

#include <stdio.h>

//求完数，即A%B=0
int main(void)
{
	//第一步先求出这个数的因子
	int n;
	int arr[] = {0};
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		int temp = 0;//temp为各因子之和
		int k = 0;
		//内循环求因子
		for (int j = 1; j<i; j++)
		{
			if (i % j == 0)
			{
				temp += j;
				arr[k]=j;//求出每个因子放在数组arr[]中
				k++;
			}
		}
		//判断是否为完数
		if(temp==i)
		{
			printf("%d its factors are", i);
			for (int x = 0; x< k; x++)
			{
				printf(" %d", arr[x]);
			}
			printf("\n");
		}

	}
	return 0;
}

答案为0 不知道是为啥啊

#include<stdio.h>
int main()
{
    int a,b,count=0,N;
    scanf("%d",&N);
    for(a=2;a<=N;a++){
        for(b=1;b<=(a/2);b++)
        {
        if(a%b==0)
        count+=b;
    }
    if(count==a)
    {
        printf("%d its factors are",a);
    for(b=1;b<=(a/2);b++){
        if(a%b==0)
        printf(" %d",b);
    }
    printf("\n");
       }
       count=0;
          }
}