| Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
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 (-10 ≤ a ≤ 10) 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
Source
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++);
else
printf("Case %d: divisible\n",k++);
}
return 0;
}