HDOJ 1212 Big Number(大数版同余定理)

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5916    Accepted Submission(s): 4135

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
   
   
   
   

 

乍一看还以为是大数相减，通过不断相减得到小于n的值，即余数。准备敲时，感觉会超时，放弃了。

 

正解则是使用同余定理不断对s取模。

 

这一题用了以下两个同余定理公式：

              (A + B) mod M = ( A mod M + B mod M ) mod M
                                                                    

              (A * B) mod M = ((A mod M) *( B mod M)) mod M  

代码如下：

 

#include<cstdio>
#include<cstring>
int main()
{
	int n,len,i,ans;
	char str[1010];
	while(scanf("%s%d",str,&n)!=EOF)
	{
		len=strlen(str);
		ans=0;
		for(i=0;i<len;i++)
		{
			ans=ans*10+(str[i]-'0');
			ans=ans%n;
		}
		printf("%d\n",ans);
	}
	return 0;
}