1 scanf("%s%d",a,&b);
2 int len=strlen(a);
3 int ans=0;
4 for(int i=st;i<len;i++)
5 ans=(int)(((long long)ans*10+a[i]-‘0‘)%b);
6 printf("%d\n",ans);
Description
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
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include<cmath>
6 using namespace std;
7 #define maxn 10000000
8 int t,b;
9 char a[205];
10 int main()
11 {
12 scanf("%d",&t);
13 int cnt=0;
14 while(t--)
15 {
16 scanf("%s%d",a,&b);
17 int len=strlen(a);
18 int ans=0;
19 int st=0;
20 printf("Case %d: ",++cnt);
21 if(a[0]==‘-‘)
22 st=1;
23 for(int i=st;i<len;i++)
24 ans=(int)(((long long)ans*10+a[i]-‘0‘)%b);
25 if(ans==0)
26 printf("divisible\n");
27 else
28 printf("not divisible\n");
29
30 }
31
32 return 0;
33 }
34
35 //6
36 //
37 //101 101
38 //
39 //0 67
40 //
41 //-101 101
42 //
43 //7678123668327637674887634 101
44 //
45 //11010000000000000000 256
46 //
47 //-202202202202000202202202 -101