LightOJ 1214(Large Division)

Posted 2020-10-28 ACRykl

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

 

分析：大数取余。

#include<cstdio>
char s[300];
int main()
{
    int T,cas=0;long long b;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%lld",s,&b);
        if(b<0) b=-b;
        int i=0;
        if(s[i]==‘-‘) i++;
        long long res=0;
        while(s[i])
            res=(res*10+(s[i++]-‘0‘))%b;
        if(res==0) printf("Case %d: divisible\n",++cas);
        else printf("Case %d: not divisible\n",++cas);
    }
    return 0;
}

View Code