高精度运算——A mod B

Big Number

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

2 312 7152455856554521 3250

Sample Output

251521

 

仍然是小学知识给的启示。列出一个除法的竖式，从最高位开始除，余数乘10加下一位上的数，再除。循环进行。

所以代码非常简单。

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1. #include<stdio.h>  
2.   
3. #include<string.h>  
4.   
5. int main()  
6.   
7. {  
8.   
9.  intn,sum,i,k;  
10.   
11.  char s[1001];  
12.   
13.  while(scanf("%s%d",s,&n)!=EOF)  
14.   
15.  {  
16.   
17.  k=strlen(s);  
18.   
19.  sum=0;  
20.   
21.  for(i=0;i<k;i++)  
22.   
23.   {  
24.   
25.   sum=sum*10+s[i]-'0';  
26.   
27.   sum=sum%n;  
28.   
29.   }  
30.   
31.  printf("%d\n",sum);  
32.   
33.  }  
34.   
35.  return 0;  
36.   
37. }