[FJOI2007]轮状病毒

嘟嘟嘟

这是一篇假题解

刚开始这道题我觉得是矩阵树定理，然而好像还得用高斯消元求行列式，不太会呀……然后想了半天dp式也没想出来，看了题解还是不太懂，最后lba，qmcp两人告诉了我一个玄学的方法。

首先f[1] = 1, f[2] = 3，然后像斐波那契一样递推直到第n项，如果n为奇数，就输出f[n] * f[n]，偶数就输出f[n] * f[n] - 4……然而并不会证明。最后提醒一下，别忘用高精度。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter printf("
")
#define space printf(" ")
#define Mem(a) memset(a, 0, sizeof(a))
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const int eps = 1e-8;
const int maxn = 105;
const int max_size = 1e3 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch))
    {
        ans = ans * 10 + ch - '0'; ch = getchar();
    }
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

struct Big
{
    int len, a[max_size];
    Big() {Mem(a); len = 0;}
    void init(int x)
    {
        for(int i = x; i > 0; i /= 10) a[++len] = i % 10;
    }
    Big operator + (const Big& b)const
    {
        Big c;
        int l = max(len, b.len); l++; 
        for(int i = 1; i <= l; ++i)
        {
            c.a[i] += (a[i] + b.a[i]) % 10;
            c.a[i + 1] += (a[i] + b.a[i]) / 10;
        }
        while(l && !c.a[l]) l--;
        c.len = l;
        return c;
    }
    Big operator * (const Big& b)const
    {
        Big c;
        for(int i = 1; i <= len; ++i)
            for(int j = 1; j <= b.len; ++j)
            {
                c.a[i + j - 1] += a[i] * b.a[j];
                c.a[i + j] += c.a[i + j - 1] / 10;
                c.a[i + j - 1] %= 10;
            }
        int l = len + b.len;
        while(l && !c.a[l]) l--;
        c.len = l;
        return c;
    }
    Big operator - (const Big& b)
    {
        int l = max(len, b.len);
        for(int i = 1; i <= l; ++i)
        {
            if(a[i] < b.a[i]) {a[i + 1]--; a[i] += 10;}
            a[i] -= b.a[i];
        }
        while(l && !a[l]) l--;
        len = l;
        return *this;
    }
    void out()
    {
        for(int i = len; i > 0; --i) write(a[i]);
    }
};

int n;
Big f[maxn], _4, ans;

int main()
{
    n = read();
    f[2].init(3); f[1].init(1); _4.init(4);
    for(int i = 3; i <= n; ++i) f[i] = f[i - 1] + f[i - 2];
    ans = (n & 1) ? f[n] * f[n] : f[n] * f[n] - _4;
    ans.out(); enter;
    return 0;
}