完数问题

【描述】

一个数如果恰好等于它的因子之和，这个数就称为“完数”。例如，6的因子为1，2，3，而6=1+2+3，因此6是“完数”。

找出1000之内的所有完数。

【C语言】

#include<stdio.h>
#include<math.h>

int main() {
int i, j;
int t;
for (i = 1; i <= 1000; i++) {
t = 0;
for (j = 1; j <= i / 2; j++) {
if (i % j == 0)
t = t + j;
}
if (i == t) {
printf("%d", i);
}
}
return 0;
}

若按如下形式输出呢？

6 its factors are 1,2,3

【输出结果】

6 its factors are 1,2,3

28 its factors are 1,2,4,7,14

496 its factors are 1,2,4,8,16,31,62,124,248

#include<stdio.h>
#include<math.h>

int main() {
int i, j, k;
int sum;
for (i = 6; i <= 1000; i++)
{
sum = 1;
for (j = 2; j <= i / 2; j++)
{
if (i % j == 0)
sum = sum + j;
}
if (i == sum) {
printf("\n%d its factors are 1", i);
for (k = 2; k <= i / 2; k++) {
if (i % k == 0)
printf(",%d", k);
}
}
}
printf("\n");
return 0;
}