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.
OutputFor each case, print the case number first. Then print ‘divisible‘ if a is divisible by b. Otherwise print ‘not divisible‘.
Sample Input6
101 101
0 67
-101 101
7678123668327637674887634 101
11010000000000000000 256
-202202202202000202202202 -101
Sample OutputCase 1: divisible
Case 2: divisible
Case 3: divisible
Case 4: not divisible
Case 5: divisible
Case 6: divisible
分析：大数取余。
#include<cstdio> char s[300]; int main() { int T,cas=0;long long b; scanf("%d",&T); while(T--) { scanf("%s%lld",s,&b); if(b<0) b=-b; int i=0; if(s[i]==‘-‘) i++; long long res=0; while(s[i]) res=(res*10+(s[i++]-‘0‘))%b; if(res==0) printf("Case %d: divisible\n",++cas); else printf("Case %d: not divisible\n",++cas); } return 0; }
