LightOJ Large Devision (大整数取模）

最新推荐文章于 2023-04-18 00:57:19 发布

foreverzili

最新推荐文章于 2023-04-18 00:57:19 发布

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本文链接：https://blog.youkuaiyun.com/foreverzili/article/details/77535664

本文介绍了一种处理大整数除法问题的方法，通过字符串存储大整数并利用同余性质来判断两个大整数间是否存在整除关系。文章包含了一个C++实现示例，演示了如何读取大整数并进行模运算来检查是否可以被另一个数整除。

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1214 - Large Division

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

101 101

0 67

-101 101

7678123668327637674887634 101

11010000000000000000 256

-202202202202000202202202 -101

Case 1: divisible

Case 2: divisible

Case 3: divisible

Case 4: not divisible

Case 5: divisible

Case 6: divisible

思路：

* 同余：

(1). (a + b )% n = (a%n +b%n ) % n

(2) (a - b ) % n =(a%n -b%n +n)%n

(3) ab%n = (a%n)(b%n)%n;

*大整数取模：

如 1234=((1*10+2)*10+3)*10+4) 按照1.3公式每步取模。

*本题注意数据类型

代码：

#include<cstdio>  
#include<cstring>    
typedef long long LL;
     
char s[301];//字符串储存大整数
  
int main()  
{  
    int t;  
    scanf("%d", &t);    
   	for(int i=1; i<=t; i++)  
    {
	    LL ans=0;//注意数据类型
	    LL d; 
        scanf("%s%lld", s, &d);  
        int len=strlen(s);    
        for(int j=0; j<len; j++)  //可简单理解为模拟算术求余过程
        {  
            if(s[j]=='-')  continue;  
            ans=(ans*10+s[j]-'0')%d;  
        }  
        printf("Case %d: ",i);  
        if(ans) 
				printf("not divisible\n");  
        else 
			    printf("divisible\n");  
    }  
    return 0;  
}