題目：

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.

Input

Input starts with an integer?T (≤ 525), denoting the number of test cases.

Each case starts with a line containing two integers?a (-10^200?≤ a ≤ 10^200)?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

注：需要判斷輸入數(shù)字字符串a(chǎn)的首位為‘-’的情況和b為負(fù)數(shù)的情況。

代碼：

#include <iostream> #include <cstring> using namespace std; char a[100000]; int main() {long long t,i,k,b,n,h;cin>>t;h=1;while(t--){cin>>a;cin>>b;if(b<0) b=-b;n=strlen(a);if(a[0]=='-'){for(i=0;i<n-1;i++){a[i]=a[i+1];}n--;}k=0;for(i=0;i<n;i++){k=(k*10+a[i]-'0')%b;}cout<<"Case "<<h++<<": ";if(!k) cout<<"divisible"<<endl;else cout<<"not divisible"<<endl;}return 0; }

?

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