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
题意：给出t组数据，判断前者是否能被后者整除，能的话输出divisible，否则输出not divisible
思路：前者数据太大，用字符串输入，后者直接int输入即可。然后将前者每一位转换成int型，在每一位转换的时候进行取余。
但是需要注意判断两者的正负，前者若为负直接将后面每一位往前移一位，且字符串长度记得-1；后者直接变符号即可。
取余操作：
for(int i=0; i<L; i++)
{
k=(k*10+a[i]-'0')%b;
}
上述还可以写成k=(k*10%b+a[i]-'0')%b;
同余定理：
(a+b)%c=(a%c+b%c)%c
(a*b)%c=(a%c*b%c)%c
大数取余操作：
一个大数对一个数取余，可以把大数看成各位数的权值与个
位数乘积的和。
1234 = ((1 * 10 + 2) * 10 + 3) * 10 + 4
1 #include<stdio.h>
2 #include<cmath>
3 #include<string.h>
4 typedef long long ll;
5 using namespace std;
6
7 //const int N=2e200;
8 char a[20000];
9 int b;
10
11 int main()
12 {
13 int n;
14 int t;
15 scanf("%d",&t);
16 int tt=1;
17 while(t--)
18 {
19 scanf("%s %d",a,&b);
20 int L=strlen(a);
21 if(a[0]=='-')//判断前者字符串是否为负
22 {
23 for(int i=1; i<L; i++)
24 {
25 a[i-1]=a[i];
26 }
27 // strlen(a)--;
28 L--;//这里注意一下
29 }
30 if(b<0)//判断后者int是否为负
31 b=-b;
32
33 ll k=0;//这里要写成ll，不然WA
34 for(int i=0; i<L; i++)
35 {
36 k=(k*10+a[i]-'0')%b;
37 }
38 if(k==0)
39 printf("Case %d: divisible\n",tt++);
40 else
41 printf("Case %d: not divisible\n",tt++);
42 }
43 return 0;
44 }