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
![](http://acm.hdu.edu.cn/data/images/C177-1003-1.gif)
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
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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; }
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