Light oj--1214(大数整除)

Description

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

代码如下:

#include<stdio.h>#include<string.h>#include<cmath>int main(){int t,cas,l,b,p;cas=1;char a[250];long long m;scanf("%d",&t);while(t--){scanf("%s%d",a,&b);        printf("Case %d: ",cas++);        p=0;        if(b<0)b=-b;        if(a[0]=='-'){        p=1;        }        l=strlen(a);        m=0;        for(int i=p;i<l;i++){        m=m*10+a[i]-'0';        if(m>=b)        {        m=m%b;        }        }        if(m==0)        printf("divisible\n");        else printf("not divisible\n");}return 0;}