Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5988    Accepted Submission(s): 4184

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

C语言程序代码

这是个大数题，要用数组。有数学上的一些规律即：（a+b）%c=(a%c+b%c)%c;a*b%c=(a%c*b%c)%c;所以根据这个就可以

   
   		将大数变小， 用数组逐位取余即可，避免计算大数。 
*/
#include<stdio.h>
#include<string.h>
#include<math.h>
int main(){

   
   	char s[100001];

   
   	int a,n,m,sum,l,i;

   
   	while(scanf("%s %d",s,&n)!=EOF)

   
   	{

   
   		l=strlen(s);

   
   		sum=0;

   
   		for(i=0;i<l;i++)

   
   			sum=(sum*10+(s[i]-'0')%n)%n;

   
   			printf("%d\n",sum);

   
   	}

   
   	return 0;
}