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

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Description

Given two integers, a and b, you should checkwhethera is divisible by b or not. We know that an integerais divisible by an integer b if and only if there exists an integercsuch 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) andb (|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'ifa 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

//Memory: 1508 KB Time: 4 MS//Language: C++ Result: Accepted#include <iostream>#include <string>using namespace std;int main(){    string num;    string::iterator it;    int T, ca;    long long carry, b; // !!!!long long才行！！！！    while(cin >> T)    {        ca = 1;        while(T--)        {            cin >> num >> b;            it = num.begin();            carry = 0;            if(num[0] == '-') it++;            for(; it != num.end(); it++)            {                carry = carry * 10 + *it - '0';                carry %= b;            }            if(!carry) cout << "Case" << " " << ca++ << ": divisible" << endl;            else cout << "Case" << " " << ca++ << ": not divisible" << endl;        }    }    return 0;}