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 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

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 #include #include #include 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<