| Time Limit: 1 second(s) | Memory Limit: 32 MB |
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 |
Output for Sample Input |
|
6 101 101 0 67 -101 101 7678123668327637674887634 101 11010000000000000000 256 -202202202202000202202202 -101 |
Case 1: divisible Case 2: divisible Case 3: divisible Case 4: not divisible Case 5: divisible Case 6: divisible |
sum忘记初始化错了好久，还是同余定理
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define INF 0x3f3f3f
#define LL long long
#define MAX 2000002
using namespace std;
int main()
{
int t,n,j,i;
int k=1;
char s[MAX];
scanf("%d",&t);
while(t--)
{
memset(s,0,sizeof(s));
scanf("%s%d",s,&n);
printf("Case %d: ",k++);
if(n<0)
n=-n;
int len=strlen(s);
LL sum=0;
for(j=0;j<len;j++)
{
if(s[j]=='-')
continue;
sum=sum*10+(s[j]-'0');
sum=sum%n;
}
if(sum==0)
printf("divisible
");
else
printf("not divisible
");
}
return 0;
}