关于"完数"的
找出1000以内的所有"完数". 请帮我讲解一下怎么编的呀!我看潭老先生的书,上面关于完数的看的不是太明白.
下面是潭老先生书中写的方法 为什么他的K只取到10 是1000以内的"完数"最多只能由10个因子构成吗?谢谢大家帮忙!

#define M 1000

#include<stdio.h>

void main()
{
int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10;
int i,a,n,s;
for(a=2;a<=M;a++)
{
n=0;
s=a;
for(i=1;i<a;i++)
if(a%i==0)
{
n++;
s=s-i;
switch(n)
{
case 1:
k1=i;break;
case 2:
k2=i;break;
case 3:
k3=i;break;
case 4:
k4=i;break;
case 5:
k5=i;break;
case 6:
k6=i;break;
case 7:
k7=i;break;
case 8:
k8=i;break;
case 9:
k9=i;break;
case 10:
k10=i;break;
}
}
if(s==0)
{
printf("%d,its factor are",a);
if(n>1) printf("%d,%d",k1,k2);
if(n>2) printf(",%d",k3);
if(n>3) printf(",%d",k4);
if(n>4) printf(",%d",k5);
if(n>5) printf(",%d",k6);
if(n>6) printf(",%d",k7);
if(n>7) printf(",%d",k8);
if(n>8) printf(",%d",k9);
if(n>9) printf(",%d",k10);
printf("\n");
}

}

}

給你看看這個....哈哈...你更該佩服了...

#include "stdio.h"
#include "conio.h"

main()
{
int a,b,c,d,e,f,g;
FILE *fp;
for(a=1;a<=30;a++){
for(b=1;b<=30;b++) {
for(c=1;c<=30;c++){
for(d=1;d<=30;d++){
for(e=1;e<=30;e++){
for(f=1;f<=30;f++){
for(g=1;g<=30;g++){
if(a==b)break;
else if(b==c)break;
else if(c==d)break;
else if(d==e)break;
else if(e==f)break;
else if(f==g)break;
else if(g==a)break;
else if(a+b+c+d+e+f+g==114)
{
if((fp=fopen("1.txt","w"))==NULL)
{
printf("can not open file.\n");
exit(1);
}
fprintf(fp,"%3d%3d%3d%3d%3d%3d%3d\n",a,b,c,d,e,f,g);
fclose(fp);
}
}
}
}
}
}
}
}
getch();
}

‘完数’概念还没弄明白啊