HDU-1212-Big Number【大数】

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本文针对ACM竞赛中常见的大数取模问题进行解析，介绍了一种高效算法实现方式，该方法能够在短时间内处理长度不超过1000的大数，并返回其对较小基数的取模结果。

题目链接：点击打开链接

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7421 Accepted Submission(s): 5135

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

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
char s[1010];
int main()
{
	while(~scanf("%s %d",s,&n))
	{
		int a=0;
		for(int i=0;s[i]!='\0';i++)
			a=(a*10+s[i]-'0')%n;
		printf("%d\n",a);
	}
	return 0;
}