同余定理之杭电1212_Big Number

原创于 2016-07-24 11:48:11 发布 · 639 阅读

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本文针对ACM竞赛中常见的大数取模问题进行了详细的解析，并提供了一个简洁高效的解决方案。通过逐位处理大数，实现了快速计算任意长度的大数A对小数B取模的结果。

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

AC代码如下：

#include "iostream"
using namespace std;

int main(int argc, char* argv[])
{
	char a[1001];
	int i,b,s,sum;
	while(cin>>a>>b)
	{
		s=strlen(a);
		sum=0;
		for (i=0;i<s;i++)
		{
			sum=(sum*10+a[i]-48)%b;
		}
		cout<<sum<<endl;
	}
	return 0;
}