Large Division LightOJ

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

题意：给出两个数问第二个数能否被第一个整除。

代码：

#include<stdio.h>#include<string.h>#define N 500int main(){char s[N];long long i,j,k,m,n,t,T,p=0;scanf("%lld",&T);while(T--){p++;scanf("%s%lld",s,&n);int l=strlen(s);k=0;t=0;if(n<0)n=-n;if(s[0]=='-'){for(i=1;i<l;i++){m=s[i]-'0';k=(m+t)%n;t=k*10;}}else{for(i=0;i<l;i++){m=s[i]-'0';k=(m+t)%n;t=k*10;}}//printf("%d\n",k);if(k==0){printf("Case %d: divisible\n",p);}else{printf("Case %d: not divisible\n",p);}}return 0;}