bzoj1002 分类： bzoj 2015-06-27 21:39 24人阅读 评论(0) 收藏

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DP , fi=3∗fi−1−2∗fi−2+2。我并不会证明呢。。。

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#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<cstring>
#include<map>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<iostream>
#include<algorithm>

const int maxl = 30, maxn = 105, Base = 1e4;

struct BigNum
{
    int num[maxl], l;
#define adj(x) ((-num[x])/Base + 1)

    void digit(int t)
    {
        if(num[t] < 0)
            num[t+1] -= adj(t), num[t] += adj(t)*Base;
        if(num[t] > Base)
            num[t+1] += num[t]/Base, num[t] %= Base;

    }
    BigNum operator *(const int b)
    {
        BigNum ret = (*this);
        if(!l || !b){ret.l = 0; return ret;}
        for(int i = 1; i <= l; i++) ret.num[i] *= b;
        for(int i = 1; i <= l; i++) ret.digit(i);
        while(ret.num[ret.l+1]) ++ret.l, ret.digit(ret.l);   
        return ret;     
    }
    void operator +=(const BigNum add)
    {
        l = std::max(l, add.l);
        for(int i = 1; i <= l; i++) num[i] += add.num[i];
        for(int i = 1; i <= l; i++) digit(i);
        if(num[l+1]) l++;   
    }
    void operator +=(const int add)
    {
        num[1] += add;
        for(int i = 1; i <= l; i++)
        {
            if(num[i] >= Base)
                digit(i);
            else
                break;  
        }
        if(num[l+1]) l++;   
    }
    void operator -=(const BigNum mus)
    {
        int nl = 0;
        l = std::max(l, mus.l);
        for(int i = 1; i <= l; i++) num[i] -= mus.num[i];
        for(int i = 1; i <= l; i++) {digit(i); if(num[i]) nl = i;}
        l = nl;
    }
    void prt()
    {
        printf("%d",num[l]);
        for(int i = l-1; i >= 1; i--)
            printf("%04d", num[i]);
    }
}emp, dp[maxn];

int n;

int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj1002.in","r",stdin);
    freopen("bzoj1002.out","w",stdout);
#endif

    std::cin >> n;

    dp[1].num[1] = dp[1].l = dp[2].l = 1, dp[2].num[1] = 5;
    for(int i = 3; i <= n; i++)
        dp[i] += dp[i-1]*3, dp[i] -= dp[i-2], dp[i] += 2;

    dp[n].prt();

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;
}

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