HDU 2802 F(N)（简单题，找循环解）

题目链接
F(N)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4579 Accepted Submission(s): 1610

Problem Description

Giving the N, can you tell me the answer of F(N)?

Input
Each test case contains a single integer N(1<=N<=10^9). The input is terminated by a set starting with N = 0. This set should not be processed.

Output
For each test case, output on a line the value of the F(N)%2009.

Sample Input
1
2
3
0

Sample Output
1
7
20

Source
HDU 2009-4 Programming Contest

/*
这类题一般有规律，比如是循环，可以化简成等差、等比数列。
此题用程序跑出循环节，在对n取run的模就可以了。
跑循环节的时候也要取模%2009.
*/
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long  LL;
using namespace std;
LL f[1000000];
int run;//循环节
int n;
void init()//找循环
{
    f[1]=1;
    f[2]=7;
    for(int i=3;;i++)
    {
        f[i]=(f[i-2]+3*i*i-3*i+1)%2009;
        if(f[i]==f[1])
        {
            bool fg=true;
            int k=(i-1)*2;
            int pos=1;
            for(;i<=k;i++)
            {
                f[i]=(f[i-2]+3*i*i-3*i+1)%2009;
                if(f[i]!=f[pos])
                {
                    fg=false;
                    break;
                }
                pos++;
            }
            if(fg)
            {
                run=k/2;
                break;
            }
        }
    }
}
int main ()
{
    init();
    while(scanf("%d",&n),n)
    {
        printf("%lld
",f[n%run]);

    }
    return 0;
}