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
题意就是让你判断一个数能否被另外一个数整除，如果能就输出divisible，否者就输出not divisible。一种高精度算法，按照题意输出就好。
#include<stdio.h>
#include<string.h>
int main()
{
long long int k=1,n;
scanf("%lld",&n);
while(n--)
{
char s[1010];
long long int sum=0,m;
scanf("%s",s);
scanf("%lld",&m);
int l=strlen(s);
for(int i=0;i<l;i++)
{
if(s[i]=='-')
continue;
sum=(sum*10+s[i]-'0')%m;
}
if(sum==0)
printf("Case %lld: divisible\n",k++);
else
printf("Case %lld: not divisible\n",k++);
}
return 0;
}