题目：
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.
输入：
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这里是10的200次方) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.
输出：
For each case, print the case number first. Then print 'divisible' if a is divisible by b. Otherwise print 'not divisible'.
样例输入：
6
101 101
0 67
-101 101
7678123668327637674887634 101
11010000000000000000 256
-202202202202000202202202 -101
样例输出：
Case 1: divisible
Case 2: divisible
Case 3: divisible
Case 4: not divisible
Case 5: divisible
Case 6: divisible
这个题a的范围是10的20次方，所以用long long 的范围也不够，只能用数组求数。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define ll long long
#define maxn 101000
int main()
{
int t;
ll b,c;
int ans=1;
char s[maxn];
scanf("%d",&t);
while(t--)
{
ll num=0;
scanf("%s%lld",s,&b);
int len=strlen(s);
// if(b<0)//这里不写也AC了
// b=-b;
for(int i=0;i<len;i++)
{
if(s[i]=='-')
continue;
num=(num*10+s[i]-'0')%b;//这里要好好解释一下，一开始的时候我想的是把num这个数求出来就好了，但是忘了我们之所以要用数组就是因为long long的范围不够，所以num的总数是求不出来的，所以就出现了这个格式，拿最后一个例子来说，-202202202202000202202202，第一个202可以整除101，第二个也能整除101.......，所以最后这个大数一定能整除101，最后num的值就是0
}
printf("Case %d: ",ans++);
if(num==0)
printf("divisible\n");
else
printf("not divisible\n");
}
return 0;
}