N！Again HDU - 2674（思维）

原创于 2020-05-28 21:23:04 发布 · 250 阅读

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本文介绍了一种快速计算N的阶乘并对2009取余的算法，通过预先计算并存储小范围内的阶乘值，对于任意输入N（0<=N<=10^9），能够迅速给出N!mod2009的结果，特别针对N大于等于2009的情况进行了优化。

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

24
120

题意：

求n的阶乘！！

思路：

如果输入的n大于等于2009，那么最终阶乘取余2009等于0，所以这里可以特判，然后打表就可！

代码：

#include<stdio.h>
#include<math.h>
int a[2100];
int main()
{
    long long n,i;
    a[0]=a[1]=1;
    for(i=2;i<=2009;i++)
        a[i]=a[i-1]*i%2009;
    while(~scanf("%lld",&n))
    {
        if(n>=2009)
            printf("0\n");
        else
            printf("%d\n",a[n]);
    }
    return 0;
}