HDU1212 Big Number 解题报告

最新推荐文章于 2021-04-18 09:48:20 发布

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最新推荐文章于 2021-04-18 09:48:20 发布

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分类专栏：数论文章标签： HDU

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本文介绍了一种解决大数求模问题的有效算法，通过逐位处理大数的每一位来简化计算过程，确保即使在大数面前也能快速准确地得出AmodB的结果。

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Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3830 Accepted Submission(s): 2635

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

Author

Ignatius.L

Source

杭电ACM省赛集训队选拔赛之热身赛

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Eddy

#include<iostream>
#include<string>
using namespace std;
#define nn 1100
int main()
{
	int b;
	char a[nn];
	while(~scanf("%s%d",a,&b))
	{
		int i,j,s=0;
		int n=strlen(a);
		for(i=0;i<n;i++)
		{
			s=s*10+a[i]-'0';
			s=s%b;
			//头个数对b求余,
			//取的的余数再与下一个数组成十进制两位数对b求余
			//以此类推到最后的结果就是大数对小数求余	
		}
		printf("%d\n",s);
	}
	return 0;
}