题解报告：hdu 1212 Big Number（大数取模+同余定理）

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：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 string str;int mod, ans;
 int main() {
     while(cin >> str >> mod) {
         ans = ;
         for(int i = ; str[i]; ++i) ans = (ans *  + (str[i] - '')) % mod;
         cout << ans << endl;
     }
     return ;
 }