BZOJ1002:[FJOI2007]轮状病毒(找规律,递推)

Description

　　轮状病毒有很多变种，所有轮状病毒的变种都是从一个轮状基产生的。一个N轮状基由圆环上N个不同的基原子
和圆心处一个核原子构成的，2个原子之间的边表示这2个原子之间的信息通道。如下图所示

　　N轮状病毒的产生规律是在一个N轮状基中删去若干条边，使得各原子之间有唯一的信息通道，例如共有16个不
同的3轮状病毒，如下图所示

　　现给定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

　　计算出的不同的n轮状病毒数输出

Sample Input

3

Sample Output

16

Solution

数论推个P，打表找规律
emmm听说这个题要用矩阵树定理……我不会啊……
然后我就去找题解想学一下……然后就看到了一篇找规律的题解……
f[1]=1,f[2]=3,答案就是斐波那契数列的第n项的平方，如果n是偶数还要再减4
高精度都懒得打了……直接搬了一个
啥？矩阵树？以后再说吧

Code

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

 const int MAXN = ;

 struct bign
 {
     int len, s[MAXN];
     bign ()
     {
         memset(s, , sizeof(s));
         len = ;
     }
     bign (int num) { *this = num; }
     bign (const char *num) { *this = num; }
     bign operator = (const int num)
     {
         char s[MAXN];
         sprintf(s, "%d", num);
         *this = s;
         return *this;
     }
     bign operator = (const char *num)
     {
         for(int i = ; num[i] == ''; num++) ;  //去前导0
         len = strlen(num);
         for(int i = ; i < len; i++) s[i] = num[len-i-] - '';
         return *this;
     }
     bign operator + (const bign &b) const //+
     {
         bign c;
         c.len = ;
         for(int i = , g = ; g || i < max(len, b.len); i++)
         {
             int x = g;
             if(i < len) x += s[i];
             if(i < b.len) x += b.s[i];
             c.s[c.len++] = x % ;
             g = x / ;
         }
         return c;
     }
     bign operator += (const bign &b)
     {
         *this = *this + b;
         return *this;
     }
     void clean()
     {
         while(len >  && !s[len-]) len--;
     }
     bign operator * (const bign &b) //*
     {
         bign c;
         c.len = len + b.len;
         for(int i = ; i < len; i++)
         {
             for(int j = ; j < b.len; j++)
             {
                 c.s[i+j] += s[i] * b.s[j];
             }
         }
         for(int i = ; i < c.len; i++)
         {
             c.s[i+] += c.s[i]/;
             c.s[i] %= ;
         }
         c.clean();
         return c;
     }
     bign operator *= (const bign &b)
     {
         *this = *this * b;
         return *this;
     }
     bign operator - (const bign &b)
     {
         bign c;
         c.len = ;
         for(int i = , g = ; i < len; i++)
         {
             int x = s[i] - g;
             if(i < b.len) x -= b.s[i];
             if(x >= ) g = ;
             else
             {
                 g = ;
                 x += ;
             }
             c.s[c.len++] = x;
         }
         c.clean();
         return c;
     }
     bign operator -= (const bign &b)
     {
         *this = *this - b;
         return *this;
     }
     bign operator / (const bign &b)
     {
         bign c, f = ;
         for(int i = len-; i >= ; i--)
         {
             f = f*;
             f.s[] = s[i];
             while(f >= b)
             {
                 f -= b;
                 c.s[i]++;
             }
         }
         c.len = len;
         c.clean();
         return c;
     }
     bign operator /= (const bign &b)
     {
         *this  = *this / b;
         return *this;
     }
     bign operator % (const bign &b)
     {
         bign r = *this / b;
         r = *this - r*b;
         return r;
     }
     bign operator %= (const bign &b)
     {
         *this = *this % b;
         return *this;
     }
     bool operator < (const bign &b)
     {
         if(len != b.len) return len < b.len;
         for(int i = len-; i >= ; i--)
         {
             if(s[i] != b.s[i]) return s[i] < b.s[i];
         }
         return false;
     }
     bool operator > (const bign &b)
     {
         if(len != b.len) return len > b.len;
         for(int i = len-; i >= ; i--)
         {
             if(s[i] != b.s[i]) return s[i] > b.s[i];
         }
         return false;
     }
     bool operator == (const bign &b)
     {
         return !(*this > b) && !(*this < b);
     }
     bool operator != (const bign &b)
     {
         return !(*this == b);
     }
     bool operator <= (const bign &b)
     {
         return *this < b || *this == b;
     }
     bool operator >= (const bign &b)
     {
         return *this > b || *this == b;
     }
     string str() const
     {
         string res = "";
         for(int i = ; i < len; i++) res = char(s[i]+'') + res;
         return res;
     }
 };

 istream& operator >> (istream &in, bign &x)
 {
     string s;
     in >> s;
     x = s.c_str();
     return in;
 }

 ostream& operator << (ostream &out, const bign &x)
 {
     out << x.str();
     return out;
 }

 bign a,b;
 int n;

 int main()
 {
     scanf("%d",&n);
     if (n==){printf(""); return ;}
     if (n==){printf(""); return ;}
     a=,b=;
     for (int i=;i<=n;++i)
     {
         a=a+b;
         swap(a,b);
     }
     if (n%==) b=b*b;
     else b=b*b-;
     cout<<b;
 }