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  • hdu 2674(余数性质)

    N!Again

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4728    Accepted Submission(s): 2490


    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
     
    大于2009的直接输出0,小于的先打表。
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    using namespace std;
    typedef long long LL;
    const int N = 2009;
    short res[N];
    
    void init(){
        res[0] = 1;
        res[1] = 1;
        for(int i=2;i<=N;i++){
            res[i] = (res[i-1]*i)%2009;
        }
    }
    int main()
    {
        int n;
        init();
        while(~scanf("%d",&n)){
            if(n<=2009)
            printf("%d
    ",res[n]);
            else printf("0
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/liyinggang/p/5526996.html
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