Large Division
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit
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\n",i);
else
printf("Case %d: not divisible\n",i);
}
return 0;
}
</span>
Light Oj 1214 大数整除