HDOJ--1212(大数取余)（Big Number）

发布时间：2020-12-14 02:09:11 所属栏目：大数据来源：网络整理

导读：HDOJ--1212(大数取余)（Big Number） Big Number Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6540????Accepted Submission(s): 4533 Problem Description As we know,Big Number is always

Big Number

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6540????Accepted Submission(s): 4533

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

/*这道题不难，就是不好想，这种方法有点巧妙*/

My ? ?solution:

/*2015.12.16*/

#include<stdio.h>
#include<string.h> 
char c[2000];
int main()
{
	int i,j,k,n,m,t;
	while(scanf("%s%d",&c,&n)==2)
	{
		m=0;
		t=strlen(c);
		for(i=0;i<t;i++)//从最高位开始
		{
			m*=10;    //每进一位，余数都要乘10
			m+=c[i]-'0';//把字符转化为数字
			m%=n; 
		}
		printf("%dn",m);
	 } 
	return 0;
}

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!