BZOJ 1002 轮状病毒 矩阵树定理

题目链接：

https://www.lydsy.com/JudgeOnline/problem.php?id=1002

题目大意：

给定n(N<=100)，编程计算有多少个不同的n轮状病毒

思路：

大部分题解都是直接一个递推公式，具体得来的方法由矩阵树定理可以求得。

只是求矩阵的行列式的时候比较复杂。

具体证明过程：http://vfleaking.blog.163.com/blog/static/17480763420119685112649/

需要高精度

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);//不可再使用scanf printf
#define Max(a, b) (a) > (b) ? (a) : (b)
#define Min(a, b) (a) < (b) ? (a) : (b)
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;

typedef long long ll;
const int mod = 1e9 + 7;
const int maxn = 1000000 + 10;

#define MAXN 9999
#define MAXSIZE 1000
#define DLEN 4
//如果位数特别大，可以把数组a放大，也可以把MAXN多加几个9，同时DLEN = 9的个数 后者方法会更快
class BigNum {
private:
public:
    ll a[1000];    //可以控制大数的位数
    int len;        //大数长度

    BigNum() {len = 1;memset(a,0,sizeof(a));}   //构造函数
    BigNum(const ll);      //int 构造函数
    BigNum(const char*);    //字符串 构造函数
    BigNum(const string);   //string 构造函数
    BigNum(const BigNum &); //拷贝构造函数
    BigNum &operator=(const BigNum &);//重载赋值运算符

    friend istream& operator >> (istream&, BigNum&);//重载输入运算符
    friend ostream& operator << (ostream&, BigNum&);//重载输出运算符

    BigNum operator + (const BigNum &) const;//重载加法
    BigNum operator - (const BigNum &) const;//重载减法
    BigNum operator * (const BigNum &) const;//重载乘法
    BigNum operator / (const ll &) const;   //重载除法 int
    BigNum operator ^ (const ll &) const;   //n次方
    ll operator % (const ll &) const;      //对int 取模

    bool operator > (const BigNum & T)const; //比较大小
    bool operator > (const ll & t)const;
    bool operator < (const BigNum & T)const;
    bool operator < (const ll & t)const;
    bool operator == (const BigNum & T)const;
    bool operator == (const ll & t)const;

    void print();   //输出大数
};
BigNum::BigNum(const ll b) //int 构造函数
{
    ll c, d = b;
    len = 0;
    memset(a,0,sizeof(a));
    while(d > MAXN)
    {
        c = d - (d / (MAXN + 1)) * (MAXN + 1);
        d = d / (MAXN + 1); a[len++] = c;
    }
    a[len++] = d;
}
BigNum::BigNum(const char*s)//字符串 构造函数
{
    ll t, k, index, l;
    memset(a,0,sizeof(a));
    l = strlen(s);
    len = l / DLEN;
    if(l % DLEN)len++;
    index=0;
    for(int i = l - 1; i >= 0;i -= DLEN)
    {
        t = 0; k = i - DLEN + 1;
        if(k < 0)k = 0;
        for(int j = k; j <= i; j++)
            t = t * 10 + s[j] - '0';
        a[index++]=t;
    }
}
BigNum::BigNum(const string s)//字符串 构造函数
{
    ll t, k, index, l;
    memset(a,0,sizeof(a));
    l = s.size();
    len = l / DLEN;
    if(l % DLEN)len++;
    index=0;
    for(int i = l - 1; i >= 0;i -= DLEN)
    {
        t = 0; k = i - DLEN + 1;
        if(k < 0)k = 0;
        for(int j = k; j <= i; j++)
            t = t * 10 + s[j] - '0';
        a[index++]=t;
    }
}
BigNum::BigNum(const BigNum & T) : len(T.len)//拷贝构造函数
{
    memset(a,0,sizeof(a));
    for(int i = 0 ; i < len ; i++)a[i] = T.a[i];
}
BigNum & BigNum::operator = (const BigNum & n)//重载赋值运算符
{
    len = n.len;
    memset(a,0,sizeof(a));
    for(int i = 0 ; i < len ; i++)
        a[i] = n.a[i];
    return *this;
}
istream& operator >> (istream & in, BigNum & b)//重载输入运算符
{
    char ch[MAXSIZE * 4];
    in >> ch;
    int l = strlen(ch);
    ll count = 0, sum = 0;
    for(int i = l - 1; i >= 0;)
    {
        sum = 0;
        ll t = 1;
        for(int j = 0; j < DLEN && i >= 0; j++, i--, t *= 10)
            sum += (ch[i] - '0') * t;
        b.a[count] = sum;
        count++;
    }
    b.len = count++;
    return in;
}
ostream& operator << (ostream& out, BigNum& b)//重载输出运算符
{
    cout<<b.a[b.len - 1];
    for(int i = b.len - 2; i >= 0; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout<<b.a[i];
    }
    return out;
}
BigNum BigNum::operator + (const BigNum & T) const //重载加法
{
    BigNum t(*this);
    ll i,big;//位数
    big = T.len > len ? T.len : len;
    for(i = 0 ; i < big ; i++)
    {
        t.a[i] +=T.a[i];
        if(t.a[i] > MAXN)
        {
            t.a[i + 1]++;
            t.a[i] -= MAXN + 1;
        }
    }
    if(t.a[big] != 0) t.len = big + 1;
    else t.len = big;
    return t;
}
BigNum BigNum::operator - (const BigNum & T) const //重载减法
{
    ll i,j,big;
    bool flag;
    BigNum t1,t2;
    if(*this > T)
    {
        t1 = *this;
        t2 = T;
        flag = 0;
    }
    else
    {
        t1 = T;
        t2 = *this;
        flag = 1;
    }
    big = t1.len;
    for(i = 0 ; i < big ; i++)
    {
        if(t1.a[i] < t2.a[i])
        {
            j = i + 1;
            while(t1.a[j] == 0) j++;
            t1.a[j--]--;
            while(j > i) t1.a[j--] += MAXN;
            t1.a[i] += MAXN + 1 - t2.a[i];
        }
        else t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while(t1.a[t1.len - 1] == 0 && t1.len > 1)
    {
        t1.len--;
        big--;
    }
    if(flag)t1.a[big - 1] = 0 - t1.a[big - 1];
    return t1;
}
BigNum BigNum::operator * (const BigNum & T) const //重载乘法
{
    BigNum ret;
    ll i,j,up;
    ll temp,temp1;
    for(i = 0 ; i < len ; i++)
    {
        up = 0;
        for(j = 0 ; j < T.len ; j++)
        {
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if(temp > MAXN)
            {
                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                up = temp / (MAXN + 1);
                ret.a[i + j] = temp1;
            }
            else
            {
                up = 0;
                ret.a[i + j] = temp;
            }
        }
        if(up != 0)
            ret.a[i + j] = up;
    }
    ret.len = i + j;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
    return ret;
}
BigNum BigNum::operator / (const ll & b) const //重载除法
{
    BigNum ret;
    ll i,down = 0;
    for(i = len - 1 ; i >= 0 ; i--)
    {
        ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
    }
    ret.len = len;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
    return ret;
}
ll BigNum::operator % (const ll & b) const //重载取模
{
    ll i,d=0;
    for (i = len-1; i>=0; i--)
    {
        d = ((d * (MAXN+1))% b + a[i])% b;
    }
    return d;
}
BigNum BigNum::operator^(const ll & n) const //重载n次方
{
    BigNum t,ret(1);
    if(n < 0)exit(-1);
    if(n == 0)return 1;
    if(n == 1)return *this;
    ll m = n;
    while(m > 1)
    {
        t = *this;
        ll i;
        for(i = 1; i << 1 <= m; i <<= 1)
        {
            t = t * t;
        }
        m -= i;
        ret = ret * t;
        if(m == 1)ret = ret * (*this);
    }
    return ret;
}
bool BigNum::operator > (const BigNum & T) const //两大整数 大于号重载
{
    ll ln;
    if(len > T.len) return true;
    else if(len == T.len)
    {
        ln = len - 1;
        while(a[ln] == T.a[ln] && ln >= 0) ln--;
        if(ln >= 0 && a[ln] > T.a[ln]) return true;
        else return false;
    }
    else return false;
}
bool BigNum::operator > (const ll & t) const
{
    return *this > BigNum(t);
}
bool BigNum::operator < (const BigNum & T) const //两大整数 小于号重载
{
    return T > *this;
}
bool BigNum::operator < (const ll & t) const
{
    return BigNum(t) > *this;
}
bool BigNum::operator == (const BigNum & T) const //两大整数 等于号重载
{
    return !(T > *this) && !(*this > T);
}
bool BigNum::operator == (const ll & t) const
{
    return BigNum(t) == *this;
}
void BigNum::print()
{
    int i;
    cout << a[len - 1];
    for(i = len - 2 ; i >= 0 ; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << a[i];
    }
}

int main()
{
    int n;
    cin >> n;
    if(n == 1)puts("1");
    else if(n == 2)puts("5");
    else
    {
        BigNum a(1), b(5), c;
        for(int i = 3; i <= n; i++)
        {
            c = BigNum(3LL) * b;
            c = c - a;
            c = c + BigNum(2LL);
            a = b;
            b = c;
        }
        cout<<c<<endl;
    }
    return 0;
}