Large Division LightOJ - 1214 (大數取模)

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

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long ll;

ll pow_mod(ll a,ll b,ll c)
{
    ll ans=1;
    a%=c;
    while(b)
    {
        if(b&1) ans=ans*a%c;
        a=a*a%c;
        b>>=1;
    }
    return ans;
 } 

ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    ll r=exgcd(b,a%b,y,x);
    y-=a/b*x;
    return r;
}

ll inv(ll a,ll mod)
{
    ll x,y;
    ll d=exgcd(a,mod,x,y);
    if(d==1)    return (x%mod+mod)%mod;
    return -1;
}

int main()
{
    int t,n,m;
    scanf("%d",&t);
    char s[205];
    ll b;
    int c=t;
    while(t--)
    {
        memset(s,0,sizeof(s));
        scanf("%s %lld",s,&b);
        int i=0;
        b=b>0?b:-b;
        if(s[0]>='0'&&s[0]<='9')    i=0;
        else    i=1;
        ll ans=0;
        for(;i<strlen(s);i++)
            ans=(ans*10+s[i]-'0')%b;
        if(ans%b==0)    printf("Case %d: divisible\n",c-t);
        else            printf("Case %d: not divisible\n",c-t);

    }
    return 0;
}