Light oj 1214 Large Division（同余定理，判断是否整除）

Description

Given two integers, a and b, you should check whethera is divisible by b or not. We know that an integera is divisible by an integer b if and only if there exists an integerc 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) andb (|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' ifa 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

这道题网上很多题解都说是同余定理推出来的结论，我也不知道为什么，上网专门搜了同余定理还是不懂，不过通过这个题还是有一个很好记并且很有用的结论的：

初始化 res为0，然后：

res=(res*10+c[i]-'0')%d

若res==0则可整除这个结论

注意 i 是从0 - n不是从n - 0；

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<utility>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<math.h>
#include<iterator>
#include<stack>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const double eps=1e-8,PI=3.1415926538;
const LL MOD=1000000000+7;
const LL MAXN=4000007;

char c[301];
LL d;
LL res;
int main()
{
    int T,T1;
    scanf("%d",&T);
    T1=T;
    while(T--)
    {
        scanf("%s%lld",c,&d);
        int n=strlen(c);
        res=0;
        d=d>0?d:-d;
        for(int i=0;i<=n-1;i++)
        {
            if(c[i]=='-')continue;
            res=(res*10+c[i]-'0')%d;
        }
        res%=d;
        printf("Case %d: ",T1-T);
        if(res)printf("not divisible\n");
        else printf("divisible\n");
    }
    return 0;
}