Nanyang ACM to find the remainder

Remainder time limit: 1000 ms | Memory limit: 65535 KB Difficulty: 3 Description now give you a natural number n, its number is less than or equal to 1 million, now all you have to do is figure out the remainder after 10003. The first row has an integer m (1<=m<=8), which indicates that there are m groups of test data;
Then the M row has a natural number n per line. Output n divides the remainder after 10003 and takes one row for each output. Sample input

3
4
5
465456541

Sample output

4
5
6948

#include <iostream>
#include <stdio.h>

using namespace std;

Char a[1000005];

123% n = (((   (1%n  *  10%n + 2%n)%n*10%n)%n+3%n  )%n 
int main ()
{
	int n;< C18/>cin >> N;
	while (n--)
	{
		scanf ("%s", a);
		Long nc=0,rest;
		for (; a[nc]!= '; nc++ ')
		{
			a[nc]-= ' 0 ';
			if (NC = = 0)
			{
				rest = A[nc]% 10003
			}
			else
			{
				rest= (rest * + A[NC])%10003
			}
		cout << rest << Endl;
	}
	return 0;
}