hdu 2674 N！Again

最新推荐文章于 2024-07-21 10:00:00 发布

wchhlbt

最新推荐文章于 2024-07-21 10:00:00 发布

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分类专栏： HDU 文章标签： ACM hdu 算法

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本文介绍了一个简单的程序设计问题，即如何快速计算N!模2009的值，通过预处理2009之前的阶乘值并利用模运算特性来解决大数问题。

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N！Again

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

Total Submission(s): 1128 Accepted Submission(s): 631

Problem Description

WhereIsHeroFrom:             Zty, what are you doing ?
Zty:                                     I want to calculate N!......
WhereIsHeroFrom:             So easy! How big N is ?
Zty:                                    1 <=N <=1000000000000000000000000000000000000000000000…
WhereIsHeroFrom:             Oh! You must be crazy! Are you Fa Shao?
Zty:                                     No. I haven's finished my saying. I just said I want to calculate N! mod 2009

Hint : 0! = 1, N! = N*(N-1)!

Input

Each line will contain one integer N(0 <= N<=10^9). Process to end of file.

Output

For each case, output N! mod 2009

Sample Input

4 5

Sample Output

24120

解题思路：水题，对于2009以前的部分打表，之后都是0了。

#include <iostream>

using namespace std;
int a[3000];
int main()
{
    int n,i;
    a[0] = 1;
    for(i=1;i<=2009;i++)
    {
        a[i] = a[i-1]*i%2009;
    }
    while(cin>>n)
    {
        if(n<=2009)
        cout << a[n]<<endl;
        else
            cout << '0'<<endl;

    }
    //cout << "Hello world!" << endl;
    return 0;
}