HDU 2802 F(N)（简单题，找循环解）

F(N)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2037    Accepted Submission(s): 677

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

 

 

Recommend

lcy

 

 

 

以后遇到这样的题目，第一感觉就应该去找循环解。

其实循环解也挺好找出来的。打表，程序中判断下就出来循环了。

本题的循环是4018.

至于为什么4018是循环，现在还没有想出来，感觉挺神奇的

/*
找循环周期
周期是4018，只要模4018就可以了。f[4018]=f[0]
*/

#include<stdio.h>
const int MOD=2009;
const int MAXN=10000;
int f[MAXN];
void init()
{
    f[1]=1;
    f[2]=7;
    for(int i=3;i<4018;i++)
    {
        f[i]=f[i-2]+3*i*i-3*i+1;
        f[i]%=MOD;
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    init();
    int n;
    while(scanf("%d",&n),n)
    {
        printf("%d\n",f[n%4018]);
    }
    return 0;
}