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最新推荐文章于 2019-08-23 10:28:16 发布

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本文介绍了一种通过模拟除法从左到右判断两个大整数a和b是否能够整除的算法。输入包括多个测试用例，每个用例包含两个整数a和b，输出则是判断a是否能被b整除的结果。

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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

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：问a是否能被b整除。

思路：模拟除法，从左到右

Code：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int T;
    scanf("%d",&T);
    for(int c = 1;c <= T;c++)
    {
        char s[205];
        long long n,res=0;
        int i=0;
        scanf("%s%lld",s,&n);
        if(s[0]=='-') i++;
        for(;s[i]!='\0';i++)
        {
            res=(res*10+(s[i]-48)*1ll)%n;
        }
        printf("Case %d: %s\n",c,res?"not divisible":"divisible");
    }
}