hdoj 1212Big Number（大数取模）

最新推荐文章于 2021-08-19 13:51:23 发布

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#hdoj1212 #Big Number #大数取模 #大数问题

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本文探讨了模运算在编程中的应用，通过实例演示如何高效计算大整数模运算，并提供了一个简洁的代码模板，旨在简化ACM竞赛中常见的模运算问题解决过程。

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6148 Accepted Submission(s): 4295

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
模板：
 
for(i=0;i<len;i++)
{
sum=(sum*10+(s[i]-'0')%m)%m;
}
代码：
 #include<stdio.h>
#include<string.h>
int main()
{
	int i,m;
	char s[1000];
	while(scanf("%s%d",s,&m)!=EOF)
	{
		int sum=0;
		int len=strlen(s);
		for(i=0;i<len;i++)
		 {
		 	sum=(sum*10+(s[i]-'0')%m)%m;
		 }
		 printf("%d\n",sum);
    } 
	 
}