hdu 2674(N!Again)

最新推荐文章于 2024-07-21 10:00:00 发布

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本文介绍了一个高效算法，用于计算任意整数N的阶乘并对2009取余，通过分析2009的质因数分解，得出当N大于等于41时，结果恒为0的结论。

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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

题目大意：求n的阶乘，对2009取余。

思路：2009=7*7*41；

所以当n>=41的时候结果都为0；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
int main()
{
    ll n;
    while(~scanf("%lld",&n))
    {
        if(n==0)
            printf("1\n");
        else if(n>=41)
            printf("0\n");
        else
        {
            ll ans=1;
            for(ll i=2; i<=n; i++)
                ans=ans*i%2009;
            printf("%lld\n",ans%2009);
        }
    }
    return 0;
}