HDU 1212 Big Number（水题）大数取模

最新推荐文章于 2021-08-19 13:51:23 发布

Sun_AC

最新推荐文章于 2021-08-19 13:51:23 发布

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分类专栏：大数 HDU 文章标签： hdu 水题

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本文介绍了一个常见的ACM竞赛问题——大数取模，并提供了一段简洁高效的C语言代码解决方案。该问题要求计算一个大数A对另一个较小的数B取模的结果，其中A可能非常大，而B小于100000。

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Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4556 Accepted Submission(s): 3152

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：

#include<stdio.h>
#include<string.h>
int main(){
	char a[1010];
	int i,len,ans,b;
	while(~scanf("%s%d",a,&b)){
		len=strlen(a);
		ans=0;
		for(i=0;i<len;i++)
			ans=(ans*10+(a[i]-'0'))%b;
		printf("%d\n",ans);
	}
	return 0;
}