HDU 1212 Big Number(大整数取模)_字符串表示的大整数取模-优快云博客

本文链接：https://blog.youkuaiyun.com/CPOHUI/article/details/79838951

本文介绍了一种处理大数取模问题的有效算法，并通过C++代码实现了该算法。主要针对长度不超过1000的大整数A和小于100000的整数B进行A%B的运算。

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Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

As we know, Big Number isalways troublesome. But it's really important in our ACM. And today, your taskis 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 lessthan 25 lines.

Input

The input contains several test cases. Each test case consistsof two positive integers A and B. The length of A will not exceed 1000, and Bwill be smaller than 100000. Process to the end of file.

Output

For each test case, you haveto 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

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ps: 将大整数用字符串表示，逐步取模。比如 : 123 % 5, 可以用下面步骤分解

1 % 5 = (0 * 10 + 1) % 5 = (0 % 5 * 10 + 1) % 5 ;

12 % 5 = (1 * 10 + 2) % 5 = ( 1 % 5 * 10 + 2 ) % 5 ;

123 % 5 = (12 * 10 + 3） % 5 = (12 % 5 * 10+ 3) % 5 ;

#include<iostream>
#include<string>
#define ll long long
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    string num;
    int n;
    while(cin>>num>>n)
    {
        ll ans = 0;
        int len_num = num.length();
        for(int i = 0; i < len_num; i++)
        {
            ans = (ans * 10 + num[i] - '0') % n;
        }
        cout<<ans<<endl;
    }
    return 0;
}