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

 

 

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

 

 

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