题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1212

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

解题思路：这是一道大数取模的题目，运用到同余定理：（a+b）%c=(a%c+b%c)%c=（a+b%c）%c。（a*b）%c=(a%c*b%c)%c。本质是模拟做除法运算，过程中只需保留余数即可！举个例子：572%7=((((5%7==5)*10+7==57)%7==1)*10+2==12)%7==5。

AC代码：

1 #include 
     
       2 using namespace std; 3 typedef long long LL; 4 string str;int mod, ans; 5 int main() { 6     while(cin >> str >> mod) { 7         ans = 0; 8         for(int i = 0; str[i]; ++i) ans = (ans * 10 + (str[i] - '0')) % mod; 9         cout << ans << endl;10     }11     return 0;12 }