Large Division（大数，同余）

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
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

这道题也是将字符串变转换成数字边求余，以保证在范围之内。
ac代码：

#include<iostream>
#include<algorithm>

using namespace std;
string s;
long long int n,t;

int main(){
	int sum=0;
	cin>>t;
	while(t--){
		s.clear();
		cin>>s>>n;
		//if(n>0) n*=-1;
		int l=s.size();
		long long int ans=0;
		for(int i=0;i<l;i++){
			if(s[0]=='-') s[i]='0';
			
			ans=ans*10+s[i]-'0';
			ans%=n;
			//cout<<ans<<endl;
		}
		if(ans==0) cout<<"Case "<<++sum<<": divisible"<<endl;
		else cout<<"Case "<<++sum<<": not divisible"<<endl;
	}
	return 0;