L

Given two integers, a and b, you should check whether a is divisible by b or not. We know that an 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.

Each case starts with a line containing two integers a (-10²⁰⁰ ≤ a ≤ 10²⁰⁰) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.

Output

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

Sample Input

6

101 101

0 67

-101 101

7678123668327637674887634 101

11010000000000000000 256

-202202202202000202202202 -101

Sample Output

Case 1: divisible

Case 2: divisible

Case 3: divisible

Case 4: not divisible

Case 5: divisible

Case 6: divisible

题解：用一个数组来存储大数，运用同余定理

(a*b)%c=(a%c*b%c)%c;

判断最后的余数是不是为零。注意：要用到long long型

关于同余定理

AC代码

#include<stdio.h>
#include<string.h>

int main()
{
    int n;
    char a[230];
    int b, num = 0;
    long long sum;
    scanf("%d", &n);
    while(n--)
    {
        sum = 0;
        scanf("%s %d", a, &b);
        int len = strlen(a);
        for(int i = 0; i < len; i++)
        {
            if(a[i] != '-')
            {
                sum = (sum*10 + a[i] - '0') % b;
            }
        }
        num++;
        if(sum == 0)
            printf("Case %d: divisible
", num);
        else
            printf("Case %d: not divisible
", num);
    }

    return 0;
}