HDU 2802 F(N) 数论+打表

题目大意：f[n]-n^3=f[n-2]-(n-1)^3 (n >=3),f[1]=1,f[2]=7,求f[n]。

题目思路：将n^3移到到等式右边化简的到：f[n]=f[n-2]+3n*(n-1)+1；

因为第n项与第n-2项有关，所以知道了f[1]与f[2]的值可以分奇偶打下表，找到循环节为4018。

#include<cstdio>
#include<stdio.h>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
#define MAX 1000005
#define mod 1000000007

using namespace std;

int v[MAX];

int main()
{
    v[0]=0;
    v[1]=1;
    v[2]=7;
    int n,ans,num;
    for(n=3;n<=4018;n++)//打表
    {

        if(n%2==0)
        {
            num=4；//从4开始
            ans=0;
            while(num<=n)
            {
                int k=(num-2)/2;
                int a=(3*num)%2009;
                int b=(num-1)%2009;
                ans=(ans+((a*b)%2009+1)%2009)%2009;
                num+=2;
            }
            ans=(ans+7)%2009;
        }

        else
        {
            num=3;//从3开始
            ans=0;
            while(num<=n)
            {
                int k=(num-2)/2;
                int a=(3*num)%2009;
                int b=(num-1)%2009;
                ans=(ans+((a*b)%2009+1)%2009)%2009;
                num+=2;
            }
            ans=(ans+1)%2009;
        }
        v[n]=ans;
    }

    while(scanf("%d",&n),n)
    {
        printf("%d\n",v[n%4018]);
    }
    return 0;
}