大数取余

(a?*?b)?%?c?=?((a?%?c)?*?(b?%?c))?%?c
(a?+?b)?%?c?=?((a?%?c)?+?(b?%?c))?%?c
10000位大的数字可以分开算：
比如：
m=123
123?=?(1*10?+?2)*10?+?3
m%n?=?123%n?=?(((1%n?*?10%n?+?2%n)%n?*?10%n)?%?n?+?3%n)%n

string?m;
long?n,?d;
int?i;
cin?>>?m?>>?n;

d=0;
for?(i?=?0;?i?<?m.size();?i++)
{
???d?=?((d?*?10)?%?n?+?m[i]?-?'0')?%?n;
???//?d*10是不超过10*n的，m[i]?-?'0'是不超过10的，除非10*n或者n+9溢出，否则整个计算过程都不会溢出
}
cout?<<?d?<<?endl;

?

描述

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.

输入

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.

输出

For each test case,you have to ouput the result of A mod B.

样例输入

2 3
12 7
152455856554521 3250

样例输出

2
5
1521

?

#include <iostream>
using namespace std;
int main()
{
 string m;
 long n,d;
 int i;
 while (cin >> m >> n)
 {
     d = 0;
  for (i = 0; i < m.size(); i++)
  {
   d = ((d * 10) % n + m[i] - '0') % n;
  }
  cout << d << endl;
 }
 return 0;
}