HDUOJ 1212 大数取余 C++版+java版

迷南。 2024-02-19 19:23 45阅读 0赞

## Big Number ##

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 10701 Accepted Submission(s): 7154**

Problem Description

As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.  
  
To make the problem easier, I promise that B will be smaller than 100000.  
  
Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

Input

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

Output

For each test case, you have to ouput the result of A mod B.

Sample Input

2 3 12 7 152455856554521 3250

Sample Output

2 5 1521

Author

Ignatius.L

Source

[杭电ACM省赛集训队选拔赛之热身赛 ][ACM_]

Recommend

Eddy

我自己写的：

题意：大数取余

用一个string 接受第一个输入，

其实就是模拟一下除法运算。

#include<iostream>
    #include<cmath>
    #include<cstdio>
    #include<cstring> 
    #include<string> 
    using namespace std;
    int main()
    {
    	string a;
    	int n[1005];
    	int mod;       //表示除数 
    	int i,j;
    	int quotient[1000];  //表示商 
    	int residue;   // 表示余数 
    	while(cin>>a>>mod)
    	{
    
    		memset(n,0,sizeof(n));
    		memset(quotient,0,sizeof(quotient));
    		int current;
    		int len=a.length();
    		for(i=0;i<len;i++)
    		{
    			
    			n[i]=a[i]-'0';
    			
    		}
    		
    
    		residue=0;
    		int pre=0;
    		for(j=0;j<i;)
    		{
    			current=residue; // 把余数赋给当前数，然后进行运算 
    			while(current<mod) // 找到一个大于除数的数 
    			{
    			    
    				current=current*10+n[j];
    				j++;
    				if(j>i)
    				{
    					residue=current/10;
    					break; 
    				}
    			}
    			if(j>i)
    			break;
    			residue=current%mod;        
    			quotient[pre]=current/mod;    //存商 
    			pre=j; 
    		}
    	    cout<<residue<<endl;
    	}
    	return 0;
    }

看完评论区别人写的：

好精彩！

数学不好的痛

### 原理： ###

### (a+b)%c = ((a%c)+(b%c))%c   (a\*b)%c = ((a%c)\*(b%c))%c ###

#include<stdio.h>
    void main()
    {
    	char x[1000];
    	int n;
    	while(scanf("%s %d",x,&n)!=EOF)
    	{
    		int i,t=0;
    		for(i=0;x[i];i++)
    		{
    			t=t*10+(x[i]-'0');
    			t%=n;
    		}
    		printf("%d\n",t);
    	}
    }

java版

简单粗暴

import java.math.BigDecimal;
    import java.util.Scanner;
    
    public class Main {
    	public static void main(String[] args) {
    		Scanner sc = new Scanner(System.in);
    		while(sc.hasNext()){
    			BigDecimal a = sc.nextBigDecimal();
    			int b = sc.nextInt();
    			System.out.println(a.remainder(new BigDecimal(b)));
    		}
    	}
    
    }

精彩，精彩

不懂的话，问我哦，乐意解答

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMzI1Njk4_size_16_color_FFFFFF_t_70][]

[ACM_]: http://acm.hdu.edu.cn/search.php?field=problem&key=%BA%BC%B5%E7ACM%CA%A1%C8%FC%BC%AF%D1%B5%B6%D3%D1%A1%B0%CE%C8%FC%D6%AE%C8%C8%C9%ED%C8%FC&source=1&searchmode=source
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQxMzI1Njk4_size_16_color_FFFFFF_t_70]: https://image.dandelioncloud.cn/pgy_files/images/2024/01/29/2149522abeec49a6b588b171521716cc.png