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
模板题，记录一下
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
char a[1000];
int main()
{
int t,n,i;
cin>>t;
int h = 0;
while(t--){
ll b,ans = 0;
cin>>a>>b;
if(b < 0){
b = -b;
}
for(i = 0; i < strlen(a); i++){
if(a[0] == '-'){
a[i] = '0';
}
ans = (ans*10+(a[i]-'0'))%b;
}
if(ans == 0){
printf("Case %d: divisible
",++h);
}
else{
printf("Case %d: not divisible
",++h);
}
}
return 0;
}