Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3256    Accepted Submission(s): 2219

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

Author

Ignatius.L

Source

​​杭电ACM省赛集训队选拔赛之热身赛 ​​

Recommend

Eddy

思路：一个数模上另一个数等于10进制数展开的模相加的模。

        如 532 mod 7 =（500%7+30%7+2%7)%7;

       当然还有a*b mod c=(a mod c+b mod c)mod c;

       如35 mod 3=((5%3)*(7%3))%3

知道这些这题就很容易了。

#include<iostream>
#include<cstring>
using namespace std;
const int mm=100020;
char s[mm];
int ss;
int main()
{
  while(cin>>s>>ss)
  {
    int len=strlen(s);
    int ans=0;
    for(int i=0;i<len;i++)
    {
      ans=ans*10+s[i]-'0';
      ans%=ss;
    }
    cout<<ans<<"\n";
  }
}