大数求模

Large Division（大数求模）

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.
给定两个整数a和b，你应该检查a是否能被b整除。我们知道整数a可以被整数b整除，如果且仅当存在整数c时，a = b * c。

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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.
输入从一个整数T(≤525),表示测试用例的数量。

每个案例始于一行包含两个整数(-10200≤≤10200)和b(| | b > 0,b适用于32位有符号整型)。数字将不包含前导零。

Output
For each case, print the case number first. Then print ‘divisible’ if a is divisible by b. Otherwise print ‘not divisible’.
对于每个案例，首先打印案例编号。如果a被b整除，则打印“divisible”，否则打印“not divisible”。

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思路：对于大数运算我们要明白int，long long是不足以存储数据的。这时候我们就要想到用字符串来表示。

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Sample Input
6

101 101

0 67

-101 101

7678123668327637674887634 101

11010000000000000000 256

-202202202202000202202202 -101

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Sample Output
Case 1: divisible

Case 2: divisible

Case 3: divisible

Case 4: not divisible

Case 5: divisible

Case 6: divisible

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#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
char a[220];
int b;          //大数求模a[n],n表示10的几次方 
int main()
{   ll n,sum,t=0,i;
    scanf("%lld",&n);
    while(n--)
    {   sum=0;
        t++;
        scanf("%s%lld",a,&b);
        b=abs(b);
        for(i=(a[0]=='-');i<strlen(a);i++)
        {
            sum*=10;
            sum+=a[i]-'0';
            sum%=b;
        }
        printf("Case %lld: %s\n",t,sum?"not divisible":"divisible");
    }
    return 0;
}