40-Topic 1064: Inverse ordinal

http://ac.jobdu.com/problem.php?pid=1064 Title Description:

Set N is a four-digit number, which is 9 times times its inverse ordinal (for example: 1234 of the inverse ordinal is 4321)
To find the value of n

Relatively simple

Topic 1064: Anti-ordinal
#include <stdio.h>
#include <iostream>
using namespace std;

int main ()
{
	int n,multiple_9;
	for (n = +; 9 * N < 9999; n++)
	{
		Multiple_9 = 9 * n;
		if (multiple_9/1000 = = n 10)//4321, 1234 highest bit comparison
			if ((multiple_9%)/100 = = (n 100)/10)//4321, 1234 second bit comparison
				I F ((multiple_9%)/= = (n% 1000)/100)//4321, 1234 Third position compare
					if (multiple_9% = = n/1000)//4321, 1234 bottom comparison
						cout << n << endl;
	}
	return 0;
}