Question

编写程序，求1~1000之间所有满足各位数字的立方和等于它本身的数。

谢谢你们

Answer 1

#include<stdio.h>
#include<math.h>
void main()
{
int i,gw,sw,bw;
for(i=1;i<1000;i++)
{
bw=i/100;
sw=(i%100)/10;
gw=i%10;
if(pow(bw,3)+pow(sw,3)+pow(gw,3)==i)
printf("%d\n",i);
}

}

Answer 2

#include <iostream>
using namespace std;
int sum(int a)//求a各位数的立方和
{
int sum=0;
int temp;
while (a)
{
temp=a%10;
a/=10;
sum+=temp*temp*temp;
}
return sum;
}
int main()
{
for (int i=1;i<=1000;++i)
{
if (i==sum(i))
{
cout <<i <<"的各位数字立方和等于它本身" <<endl;
}

}
return 0;
}

Answer 3

#include <stdio.h>
main()
{
int i,j,k,s;
for(i=1;i<=1000;i++){
j=i;s=0;
while(j){
k=j%10;
s+=k*k*k;
j/=10;
}
if(s==i)
printf("%d\n",i);

}

}

Answer 4

#include "stdio.h"
main()
{
int i,j,k,n;
for(n=100;n<1000;n++)
{
i=n/100;
j=n/10%10;
k=n%10;
if(n==i*i*i+j*j*j+k*k*k)
printf("%-5d ",n);
}
getch();
}

Answer 5

#include<iostream>
using namespace std;
int main ()
{for(int i=100;i<=999;i++)
{ int n1,n2,n3,k;
n1=i/100;
n2=(i-n1*100)/10;
n3=i-n1*100-n2*10;
k=n1*n1*n1+n2*n2*n2+n3*n3*n3;
if(i==k) {cout<<" "<<i;}
}
return 0;}