高精度运算——A mod B

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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

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。

#include<stdio.h>#include<string.h>int main(){ intn,sum,i,k; char s[1001]; while(scanf("%s%d",s,&n)!=EOF) { k=strlen(s); sum=0; for(i=0;i<k;i++)  {  sum=sum*10+s[i]-'0';  sum=sum%n;  } printf("%d\n",sum); } return 0;}

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