Light Oj 1214 大数整除

Large Division
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
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Status

Practice

LightOJ 1214
Description
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

<span style="color:#6600cc;">/*******************************************
   
     author   :   Grant Yuan
     time     :   2014.8.7
     algorithm:   大数整除
     source   :   Light Oj 1214
     explain  ：  从ans初始化为0，最低位开始ans=(ans+s[i]-'0')%b;
                  计算到最后，如果ans 等于0，可以整除；
                  否则，不可以整除；
                  
**********************************************/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define INF 0x3fffffff

using namespace std;

char s[2000002];
int b;
long long ans;
int t;

int main()
{
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {    memset(s,0,sizeof(s));
        scanf("%s%d",&s,&b);
        if(b<0) b=-b;
        int l=strlen(s);
        ans=0;
        for(int j=0;j<l;j++)
        {
            if(s[j]=='-') continue;
            ans=(ans*10+(s[j]-'0'))%b;
        }
        if(ans==0)
            printf("Case %d: divisible
",i);
        else
            printf("Case %d: not divisible
",i);
    }
    return 0;
}
</span>