HDU 2674 N！Again(规律)

N！Again

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

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

   
   
   
   

    
    
    
    24
120
   
   
   
   

 

/************************************************************************/

题意：求N! mod 2009

orz

也是醉了，刚开始还傻呆呆地找规律，利用同余一直求到了41!，发现之后的结果全是0，这时才反应过来，N! mod 2009 ，当N！能够被2009整除时，取模的结果均为0，那剩下的就是判断一下什么时候能够被2009整除

对2009分解质因数，可得2009=7*7*41，所以我们使N!有约数41就可以被2009整除了，所以才会从41！开始 mod 2009的结果均为0

因此我们只需先算好前40项mod 2009的结果就可以了，N>40的结果都是0

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 41;
const int inf = 2147483647;
const int mod = 2009;
int s[N];
int main()
{
    int n,i;
    s[0]=1;
    for(i=1;i<=40;i++)
        s[i]=(s[i-1]*i)%mod;
    while(~scanf("%d",&n))
        if(n>40)
            puts("0");
        else
            printf("%d\n",s[n]);
    return 0;
}

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