HDU1212:Big Number(大数求模)

最新推荐文章于 2021-08-19 13:51:23 发布

原创最新推荐文章于 2021-08-19 13:51:23 发布 · 431 阅读

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本文详细阐述了如何解决模运算问题，通过实例演示了一种简单有效的编程方法，适用于ACM竞赛场景，特别关注于处理大数情况。文章提供了一个包含输入输出样例的完整程序代码，帮助读者理解并实践模运算的高效处理。

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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

#include<stdio.h>
#include<string.h>
int main()
{
	char a[1000];
	int b,len;
	while(~scanf("%s%d",a,&b))
	{
		len=strlen(a);
		int mod=0;
		for(int i=0;i<len;i++)
		{
			mod*=10;
			mod+=a[i]-'0';
			mod=mod%b;
		}
		printf("%d\n",mod);
	}
	return 0;
}