[PTA] 7-45 find End Number (20 minutes)

After the number of the called number is exactly equal to the sum of factors except itself. For example: 6 = 1 + 2 + 3, where a factor of 1, 2, 6. This problem requires programming to find all the numbers between arbitrary two complete positive integers m and n.

Input format:
Input give two positive integers m and n (1 <m≤n≤10000) in a row, separated by a space.

Output format:
progressive number of each factor outputted to complete a given range of the accumulated factorization form, wherein each complete number representing the number of one row, and the format of "Number of Ends factor = factor 1 + 2 + ... + K factor", ascending order and are by factor analysis. If you do not count them within range, the output "None".

Sample input:
230

Output Sample:
6 1 + 2 + 3 =
28 = 1 + 2 + 4 + 7 + 14

#include<stdio.h>
int main()
{
	int n,m,i,j,sum,flag=0;//
	scanf("%d %d",&n,&m);
	for(i=n;i<=m;i++)
	{
		sum=1;
		if(i==1) continue;	//1不是完数 
		for(j=2;j<i;j++)
		{
			if(i%j==0)
			{
				 sum=sum+j;
			} 
		}
		if(sum==i)
		{
			printf("%d = 1",sum);
			for(j=2;j<i;j++)
			{
				if(i%j==0)
				{
					printf(" + %d",j);
				}
			}
			printf("\n");
			flag=1;
		}
	}
	if(flag==0) 
	printf("None");
	return 0;
}

Knowledge Point: flag usage