码迷,mamicode.com
首页 > 其他好文 > 详细

大数取模 模板

时间:2015-07-19 20:00:28      阅读:134      评论:0      收藏:0      [点我收藏+]

标签:

1 scanf("%s%d",a,&b);
2 int len=strlen(a);
3 int ans=0;
4 for(int i=st;i<len;i++)
5     ans=(int)(((long long)ans*10+a[i]-0)%b);
6 printf("%d\n",ans);

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 (-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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 #define maxn 10000000
 8 int t,b;
 9 char a[205];
10 int main()
11 {
12 scanf("%d",&t);
13 int cnt=0;
14 while(t--)
15 {
16     scanf("%s%d",a,&b);
17     int len=strlen(a);
18     int ans=0;
19     int st=0;
20     printf("Case %d: ",++cnt);
21     if(a[0]==-)
22         st=1;
23     for(int i=st;i<len;i++)
24     ans=(int)(((long long)ans*10+a[i]-0)%b);
25     if(ans==0)
26         printf("divisible\n");
27         else
28             printf("not divisible\n");
29 
30 }
31 
32  return 0;
33 }
34 
35 //6
36 //
37 //101 101
38 //
39 //0 67
40 //
41 //-101 101
42 //
43 //7678123668327637674887634 101
44 //
45 //11010000000000000000 256
46 //
47 //-202202202202000202202202 -101

 

大数取模 模板

标签:

原文地址:http://www.cnblogs.com/ZP-Better/p/4659156.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!