表面上看這道題是問a能不能整除b,實際上還是看二者取余余數是否爲0，屬於大數取餘的範圍
a的範圍達到10的200次方，用 long long都已經不可以，需要用字符串，而b可以用long long
題目：
F - 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
代碼：（關鍵就是那一步，研究懂就可以）
#include <stdio.h>
#include <string.h>
typedef long long ll;
int main()
{
int t,k=0,i;
scanf("%d",&t);
while(t--)
{
char s[1000];
k++;
long long b;
scanf("%s%lld",s,&b);
int len=strlen(s);
ll num=0;
for(i=0;i<len;i++)
{
if(s[i]=='-')
continue;
num=(num*10+s[i]-'0')%b;//關鍵步驟
}
if(num==0)
printf("Case %d: divisible\n",k);
else
printf("Case %d: not divisible\n",k);
}
return 0;
}