BZOJ1089:[SCOI2003]严格n元树(DP,高精度)

Description

如果一棵树的所有非叶节点都恰好有n个儿子，那么我们称它为严格n元树。如果该树中最底层的节点深度为d
（根的深度为0），那么我们称它为一棵深度为d的严格n元树。例如，深度为２的严格２元树有三个，如下图：

给出n, d，编程数出深度为d的n元树数目。

Input

仅包含两个整数n, d( 0   <   n   <   =   32,   0  < =   d  < = 16)

Output

仅包含一个数，即深度为d的n元树的数目。

Sample Input

【样例输入1】
2 2

【样例输入2】
2 3

【样例输入3】
3 5

Sample Output

【样例输出1】
3

【样例输出2】
21

【样例输出2】
58871587162270592645034001

Solution

DP方程好想= =

$f[i]$表示深度不超过$i$的树的种数

$f[i]=f[i-1]^n+1$。加1是因为儿子可能为空。

最后答案为$f[d]-f[d-1]$

需要高精度……不过懒得写了直接套了个板子QAQ

Code

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #define N (509)
 using namespace std;

 struct bign
 {
     int len, s[N];
     bign ()
     {
         memset(s, , sizeof(s));
         len = ;
     }
     bign (int num) { *this = num; }
     bign (const char *num) { *this = num; }
     bign operator = (const int num)
     {
         char s[N];
         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 Pow(bign a,int b)
 {
     bign ans=;
     for (int i=; i<=b; ++i)
         ans=ans*a;
     return ans;
 }

 bign f[],ans;
 int n,d;

 int main()
 {
     scanf("%d%d",&n,&d);
     if (!d){puts(""); return ;}
     f[]=;
     for (int i=; i<=d; ++i)
         f[i]=Pow(f[i-],n)+;
     cout<<f[d]-f[d-];
 }