Large numbers divisible

Topic: LightOJ Problem :: 1214 - Large Division

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

Meaning of the questions:

　　　　To a large number divisible by a determination whether or not a B (note that a positive and negative)

Number Theory Principle:

　　　　(a + b) % p = (a % p + b % p) % p

　　　　Congruence theorems mode: https://baike.baidu.com/item/%E5%90%8C%E4%BD%99%E5%AE%9A%E7%90%86

　　　　Congruence mold theorem: https://baike.baidu.com/item/%E5%8F%96%E6%A8%A1%E8%BF%90%E7%AE%97/10739384

Code:

　　

#include<bits/stdc++.h>
using namespace std;
int main()
{
     char a[205];
     int n;
     int b;
     cin>>n;
     for(int i=1;i<=n;i++)
     {
         cin>>a>>b;
         int len=strlen(a);
         int flag=0,ans=0;
         if(a[0]=='-')
         flag=1;
         for(int j=flag;j<len;j++)
         {
             ans=(int)(((long long)ans*10+(a[j]-'0'))%b);
        }
        printf("Case %d: ",i);
        if(ans==0)
         printf("divisible\n");
        else
         printf("not divisible\n");
     }
     return 0;
 }