L - Large Division (大数， 同余)

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 (-10 ²⁰⁰ ≤ a ≤ 10 ²⁰⁰ ) 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[];

    int b, num = ;

    long long sum;

    scanf("%d", &n);

    while(n--)

    {

        sum = ;

        scanf("%s %d", a, &b);

        int len = strlen(a);

        for(int i = ; i < len; i++)

        {

            if(a[i] != '-')

            {

                sum = (sum* + a[i] - '') % b;

            }

        }

        num++;

        if(sum == )

            printf("Case %d: divisible\n", num);

        else

            printf("Case %d: not divisible\n", num);

    }

    return ;

}