L - Large Division (大数，同余)

最新推荐文章于 2025-08-22 01:16:35 发布

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原文链接：http://www.cnblogs.com/ruruozhenhao/p/8535514.html

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#数据结构与算法

本文介绍了一种使用数组存储大数并运用同余定理判断两数是否整除的方法。通过将大数转换为字符数组处理，实现了对超出常规整型范围的大数进行除法运算的判断。

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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

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

题解：用一个数组来存储大数，运用同余定理

(a*b)%c=(a%c*b%c)%c;

判断最后的余数是不是为零。注意：要用到long long型

关于同余定理

AC代码

#include<stdio.h>
#include<string.h>

int main()
{
    int n;
    char a[230];
    int b, num = 0;
    long long sum;
    scanf("%d", &n);
    while(n--)
    {
        sum = 0;
        scanf("%s %d", a, &b);
        int len = strlen(a);
        for(int i = 0; i < len; i++)
        {
            if(a[i] != '-')
            {
                sum = (sum*10 + a[i] - '0') % b;
            }
        }
        num++;
        if(sum == 0)
            printf("Case %d: divisible\n", num);
        else
            printf("Case %d: not divisible\n", num);
    }

    return 0;
}