Big Number（大整数取模）

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4789    Accepted Submission(s): 3329

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

 

Author

Ignatius.L

 

Source

杭电ACM省赛集训队选拔赛之热身赛

意解：大整数取模运算，类似于整数除法（手算），比如512 % 10，我们手算是51 % 10 = 1，之后1 * 10 + 2 = 12，在12 % 10  =  2，就结束了；
AC代码：
      

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;typedef long long ll;int main(){    char s[1100];    int mod,len;    while(cin>>s>>mod)    {        ll sum = 0;        len = strlen(s);        for(int i = 0; i < len; i++)        {            sum = (sum * 10 + s[i] - '0') % mod;        }        cout<<sum<<endl;    }    return 0;}