Light OJ 1214-large Division "congruence theorem"

1214-large Division

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Time Limit:1 second (s)

Memory limit:32 MB

Given integers, a and b, you should check whether a is divisible by b or not. We know A integer a is divisible by an integer b if and only if there exists an integer C such that a = b * C. Input

Input starts with an integer T (≤525), denoting the number of test cases.

Starts with a line containing-integers a ( -10200≤a≤10200) and B (|b| > 0, b fits into a Signed integer). Numbers would not contain leading zeroes. Output

For each case, print the case number first. Then print ' divisible ' if a was divisible by b. Otherwise print ' not divisible '.

Sample Input	Output for Sample Input
6 101 101 0 67 -101 101 7678123668327637674887634 101 11010000000000000000 256 -202202202202000202202202-101	Case 1:divisible Case 2:divisible Case 3:divisible Case 4:not Divisible Case 5:divisible Case 6:divisible

#include <cstring>
#include <cstdio>
char a[2200000];
int b;
int main () {
	int t,t;
	scanf ("%d", &t);
	for (t = 1; t <= t;t++)
	{
		scanf ("%s", a);
		scanf ("%d", &b);
		int len = strlen (a);
		Long Long sum = 0;//int  on WA 
		..... for (int i = 0; i < len; i++)
		{
			if (a[i]== '-')	continue;
			sum = (sum*10+a[i]-' 0 ')%b;
		}
		if (sum==0)
			printf ("Case%d:divisible\n", t);
		else
			printf ("Case%d:not divisible\n", t);	
	}
	return 0;
}