Lightoj1214——Large Division(同余定理,判断是否整除)

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

Output for Sample Input

6

101 101

0 67

-101 101

7678123668327637674887634 101

11010000000000000000 256

-202202202202000202202202 -101

Case 1: divisible

Case 2: divisible

Case 3: divisible

Case 4: not divisible

Case 5: divisible

Case 6: divisible



直接套结论,如果一个数a能整除b,那么当且仅当a的各位数之和能整除b。例:判断一个数是否是3的倍数,只需判断它的各位数之和是否是3的倍数


#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<cmath>
#define MAXN 1000010
char s[210];
using namespace std;
int main()
{
    long long b,len,ans,k;
    int t,cnt=1,i,j;
    scanf("%d",&t);
    while(t--)
    {
        cin>>s>>b;
        if(b<0)
            b=-b;
        len=strlen(s);
        if(s[0]=='-')
        {
            ans=s[1]-'0';
            k=2;
        }
        else
        {
            ans=s[0]-'0';
            k=1;
        }
        ans=ans%b;
        for(i=k;i<len;++i)
            ans=(ans*10+(s[i]-'0'))%b;
        if(!ans)
            printf("Case %d: divisible\n",cnt++);
        else
            printf("Case %d: not divisible\n",cnt++);
    }
    return 0;
}



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