lightoj--1214--Large Division（大数取余）

Large Division

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

SubmitStatus

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

Source

Problem Setter: Jane Alam Jan

水题一枚

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char s[1001];int main(){int t,k=1;scanf("%d",&t);while(t--){memset(s,'\0',sizeof(s));scanf("%s",s);long long mod,temp;scanf("%lld",&mod);temp=0;for(int i=0;i<strlen(s);i++){if(s[i]=='-') continue;temp=temp*10+(s[i]-'0');temp%=mod;}if(temp%mod)printf("Case %d: not divisible\n",k++);elseprintf("Case %d: divisible\n",k++);}return 0;}