int n=2,i,j,k=0,a[100],shu=0; printf("%d以内的完数有:",N); while(n<=N) { if(ws(n)) { for(i=1;i<n;i++) if(n%i==0) a[k++]=i; printf("\n%d: %-4d=%d",shu+1,n,a[0]); for(j=1;j<k;j++) printf("+%d ",a[j]); printf(".\n"); shu++; k=0; }...