hdu 1212 Big Number(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:

 1 #include <bits/stdc++.h>
 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 }

 

posted @ 2018-02-26 16:30  霜雪千年  阅读(288)  评论(0编辑  收藏  举报