BZOJ 1002 轮状病毒

Description

给定n(N<=100)，编程计算有多少个不同的n轮状病毒。

Input

第一行有1个正整数n。

Output

将编程计算出的不同的n轮状病毒数输出

Sample Input

3

Sample Output

16

两种做法：

1.基尔霍夫矩阵+高斯消元暴搓

2.根据基尔霍夫矩阵推出推出递推公式推出f[i]=(f[i-2]*3-f[i-1]+2)。(解释见：http://vfleaking.blog.163.com/blog/static/17480763420119685112649/)

代码如下：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
using namespace std;

int n;
struct node
{
    int a[],l;
    node()
    {
        memset(a,,sizeof(a));
        l = ;
    }
    friend inline node operator *(int x,node &y)
    {
        node ret; ret.l = y.l+;
        for (int i = ;i <= y.l;++i)
        {
            ret.a[i] += y.a[i]*x;
            ret.a[i+] += ret.a[i]/;
            ret.a[i] %= ;
        }
        if (ret.a[ret.l] == ) ret.l--;
        return ret;
    }
    friend inline node operator -(node x,node y)
    {
        node z; z.l = max(x.l,y.l);
        for (int i = ;i <= z.l;++i)
        {
            z.a[i] = x.a[i]-y.a[i];
            while (z.a[i] < )
                z.a[i] += ,x.a[i+]--;
        }
        while (z.l > &&z.a[z.l] == )z.l--;
        return z;
    }
    friend inline node operator +(node &x,int y)
    {
        node ret = x;
        ret.a[] += y;
        for (int i = ;i <= ret.l;++i)
        {
            if (ret.a[i] >= )
                ret.a[i]-=,ret.a[i+]++;
            else break;
        }
        if (ret.a[ret.l+]) ret.l++;
        return ret;
    }
    inline void print()
    {
        for (int i = l;i >= ;--i)
            printf("%d",this->a[i]);
    }
}f[];

int main()
{
    freopen("1002.in","r",stdin);
    freopen("1002.out","w",stdout);
    scanf("%d ",&n);
    f[] = f[]+; f[] = f[] + ;
    for (int i = ;i <= n;++i)
    {
        f[i] = *f[i-];
        f[i] = f[i]-f[i-];
        f[i] = f[i]+;
    }
    f[n].print();
    fclose(stdin); fclose(stdout);
    return ;
}