LightOJ 1214 K - 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 $a({-10}^{200}\le a\le {10}^{200})$ 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

比较水的题目，本来想试试 python 结果发现没有😂 那就直接模拟大数取余即可，AC代码如下：

#include<bits/stdc++.h> using namespace std; typedef long long ll; char s[205]; ll t,n; int main(){ scanf("%lld",&t); for(ll i=1;i<=t;i++){ scanf("%s%lld",s,&n); int len=strlen(s); ll ans=0; for(ll i=0;i<len;i++){ if(!isdigit(s[i])) continue; ans=(ans*10+s[i]-'0')%n; } if(!ans) printf("Case %lld: divisible\n",i); else printf("Case %lld: not divisible\n",i); } }