Large Division 同余定理

最新推荐文章于 2020-12-08 12:42:57 发布

Big_Rui

最新推荐文章于 2020-12-08 12:42:57 发布

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本文介绍了一种判断大数a是否能被较小整数b整除的方法，通过逐位乘加取余运算实现，适用于a的数值范围远超常规整型变量的情况。文章提供了完整的C++代码示例。

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题意：一个非常大的数a是否能整除b

别人都说是同余，我上网看定理没研究出来啥，可能是推出来的结论，记住就好

ans=(ans*10+a[i]-'0')%b，i从0-n-1，最后结果为0就是能整除

注意：b用long long 不然会炸，不知道为什么，希望大佬看的时候评论告诉我；

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;

int main()
{
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        char a[305];
        ll b,ans=0;
        scanf("%s%lld",a,&b);
        int n=strlen(a);
        b=abs(b);
        for(int i=0;i<n;i++)
        {
            if(a[i]=='-') continue;
            ans=(ans*10+a[i]-'0')%b;
        }
        printf("Case %d: ",cas++);
        if(ans) printf("not divisible\n");
        else printf("divisible\n");
    }
}

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