标签：its put lld algorithm visible exp span print ret
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 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
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <iomanip> #include <cmath> #include <ctime> #include <map> #include <set> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>y?x:y) #define min(x,y) (x<y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define INF 0x3f3f3f3f3f #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; char a[250]; ll b,t,ans; int main() { scanf("%d",&t); int k=t; while(t--) { scanf("%s %lld",a,&b); b=abs(b); printf("Case %d: ",k-t); ans=0; for(int i=0;i<strlen(a);i++) { if(a[i]==‘-‘) continue; ans=ans*10+(a[i]-‘0‘); ans=ans%b; if(ans==0) {puts("divisible");goto end;} } puts("not divisible"); end:; } return 0; }
标签：its put lld algorithm visible exp span print ret
原文地址：http://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7236089.html
