LightOJ 1214 - Large Division【同余定理】

1214 - Large Division

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Time Limit: 1 second(s)Memory Limit: 32 MB

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

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

 

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PROBLEM SETTER: JANE ALAM JAN

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char map[10000];int main(){int t;int uu=0;scanf("%d",&t);while(t--){uu++;int nn;scanf("%s",map);int len=strlen(map);scanf("%d",&nn);long long ans=0;if(map[0]=='-'){for(int i=1;i<len;i++){ans=ans*10+(map[i]-'0');ans=ans%nn;}}else{for(int i=0;i<len;i++){ans=ans*10+(map[i]-'0');ans=ans%nn;}}printf("Case %d: ",uu);if(ans%nn==0){printf("divisible\n");}elseprintf("not divisible\n");}return 0;}