题解 | #变种水仙花#

#include <stdio.h>

int main()
{
	int sum0[5] = { 0 };
	for (int i = 10000; i <= 99999; i++)
	{
		int j = i;
		int sum = 0;
		for (int h=0; h<=3;h++)
		{
			int m = (j / pow(10, h + 1));
			int n = (j % (int)pow(10, h + 1));
			//h为控制小sum,k为控制位数相关
			//前者为万位,后者为千百个位,
			//再次时前者为千万位,后者为百个位
			sum0[h] = m * n;
			sum = sum + sum0[h];
		}
		if (sum == i)
		{
			printf("%d ", i);
		}

	}
	return 0;
}