我这里给出一种“大人的”解法，思路就是从所有可能满足条件的数中暴力搜索即10000 - 99999，每一个数字我都去求它的回文数字， 并且观察每一位是不是否不同，如果都不同，我就用回文数除以原来的数字，得出的数字是整数（确切的说是一位整数）就将其返回。

先说答案， 答案是 21978 * 4 = 87912

为了方便理解，我这里做了两次循环，其实可以放到一起, 大神请忽略～

function reversedNumber(n) {
  let r = 0;

  while (n >>> 0 !== 0) {
    r = r * 10 + (n % 10);
    n = (n / 10) >>> 0;
  }

  return r;
}

function hasDuplicated(n) {
  const  used = {};

  while (n >>> 0 !== 0) {
    n = (n / 10) >>> 0;
    if (used[n % 10]) return true;
    used[n % 10] = true;
  }

  return false;
}
function t() {
  let n = 0;
  for (let i = 10000; i < 100000; i++) {
    if(hasDuplicated(i)) continue;
    n = reversedNumber(i);
    if (n / i === (n / i) >>> 0) return `${i} * ${n / i} = ${n}`;
  }
}

console.log(t()); // 21978 * 4 = 87912

？

？

？

因为是五个不同的数字，其实循环次数可以再减少一些，就从12345-98765进行搜索

#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
	for( int i=12345; i<=98765; i++)
	{
		int a = i / 10000;
		int b = i % 10000 / 1000;
		int c = i % 10000 % 1000 / 100;
		int d = i % 10000 % 1000 % 100 / 10;
		int e = i % 10;
		if(a!=b && a!=c && a!=d && a!=e && b!=c && b!=d && b!=e && c!=d && c!=e && d!=e)
		{
		    int j = e * 10000 + d * 1000 + c * 100 + b * 10 + a;
		    if( j % i == 0 )
		    {
			cout << i << "*" << (j/i) << "=" << j <<endl;
		    }
                }
	}
	return 0;
}

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#include<stdio.h>
int main()
{
int a,b,c,d,e,f,i;
for(i=10000;i<=99999;i++)
for(f=1;f<=9;f++)
{
a=i%10;
b=i/10%10;
c=i/100%10;
d=i/1000%10;
e=i/10000;
if(if==a10000+b1000+c100+d*10+e&&f!=a&&f!=b&&f!=c&&f!=d&&f!=e)
if(a!=b&&a!=c&&a!=d&&a!=e&&b!=c&&b!=d&&b!=e&&c!=d&&c!=e&&d!=e)
printf("%d %d\n",i,f);
}
}
结果是21978 *4=87912

This issue has been automatically marked as stale because it has not had recent activity. It will be closed if no further activity occurs. Thank you for your contributions.