Large Division (大数取余，有个坑爹的地方)

Large Division

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

SubmitStatus

原题链接：点击打开链接


Description

Given two integers, a and b, you should checkwhether a is divisible by b or not. We know that an integer ais divisible by an integer b if and only if there exists an integer csuch 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 leadingzeroes.

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

这是一道很普通的大数取余的题目，可是有个地方卡了半天TT，卡住的地方就是程序注释处

1 //Memory: 1508 KB         Time: 4 MS
 2 //Language: C++         Result: Accepted
 3 
 4 #include <iostream>
 5 #include <string>
 6 using namespace std;
 7 
 8 int main()
 9 {
10     string num;
11     string::iterator it;
12     int T, ca;
13     long long carry, b; // !!!!long long才行！！！！
14     while(cin >> T)
15     {
16         ca = 1;
17         while(T--)
18         {
19             cin >> num >> b;
20             it = num.begin();
21             carry = 0;
22             if(num[0] == '-') it++;
23             for(; it != num.end(); it++)
24             {
25                 carry = carry * 10 + *it - '0';
26                 carry %= b;
27             }
28             if(!carry) cout << "Case" << " " << ca++ << ": divisible" << endl;
29             else cout << "Case" << " " << ca++ << ": not divisible" << endl;
30         }
31     }
32     return 0;
33 }

原文链接: https://www.cnblogs.com/cszlg/archive/2012/09/01/2910468.html

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