BZOJ 1089 （SCOI 2003） 严格n元树

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

　　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

———————————————————————————-

题解

转移方程比较容易思考：

    dp[i]=dp[i-1]^n+1;

主要考点就是高精的重载运算符。

代码

#include <iostream>

#include <string>

#include <cstring>

#include <cstdio>

using namespace std;

const int maxn = 10000;

struct bign{

    int d[maxn], len;

    void clean() { while(len > 1 && !d[len-1]) len--; }

    bign()          { memset(d, 0, sizeof(d)); len = 1; }

    bign(int num)   { *this = num; }

    bign(char* num) { *this = num; }

    bign operator = (const char* num){

        memset(d, 0, sizeof(d)); len = strlen(num);

        for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';

        clean();

        return *this;

    }

    bign operator = (int num){

        char s[2000]; sprintf(s, "%d", num);

        *this = s;

        return *this;

    }

    bign operator + (const bign& b){

        bign c = *this; int i;

        for (i = 0; i < b.len; i++){

            c.d[i] += b.d[i];

            if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++;

        }

        while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;

        c.len = max(len, b.len);

        if (c.d[i] && c.len <= i) c.len = i+1;

        return c;

    }

    bign operator - (const bign& b){

        bign c = *this; int i;

        for (i = 0; i < b.len; i++){

            c.d[i] -= b.d[i];

            if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;

        }

        while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;

        c.clean();

        return c;

    }

    bign operator * (const bign& b)const{

        int i, j; bign c; c.len = len + b.len;

        for(j = 0; j < b.len; j++) for(i = 0; i < len; i++)

            c.d[i+j] += d[i] * b.d[j];

        for(i = 0; i < c.len-1; i++)

            c.d[i+1] += c.d[i]/10, c.d[i] %= 10;

        c.clean();

        return c;

    }

    bign operator / (const bign& b){

        int i, j;

        bign c = *this, a = 0;

        for (i = len - 1; i >= 0; i--)

        {

            a = a*10 + d[i];

            for (j = 0; j < 10; j++) if (a < b*(j+1)) break;

            c.d[i] = j;

            a = a - b*j;

        }

        c.clean();

        return c;

    }

    bign operator % (const bign& b){

        int i, j;

        bign a = 0;

        for (i = len - 1; i >= 0; i--)

        {

            a = a*10 + d[i];

            for (j = 0; j < 10; j++) if (a < b*(j+1)) break;

            a = a - b*j;

        }

        return a;

    }

    bign operator += (const bign& b){

        *this = *this + b;

        return *this;

    }

    bool operator <(const bign& b) const{

        if(len != b.len) return len < b.len;

        for(int i = len-1; i >= 0; i--)

            if(d[i] != b.d[i]) return d[i] < b.d[i];

        return false;

    }

    bool operator >(const bign& b) const{return b < *this;}

    bool operator<=(const bign& b) const{return !(b < *this);}

    bool operator>=(const bign& b) const{return !(*this < b);}

    bool operator!=(const bign& b) const{return b < *this || *this < b;}

    bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);}

    string str() const{

        char s[maxn]={};

        for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';

        return s;

    }

};

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 f[35];

int n;

int d;

int main(){

    cin>>n>>d;

    if(n==1&&d==1) return cout<<0,0;

    if(d==0) return cout<<1,0;

    f[1]=1;

    for(int i=1;i<=d;i++) {

        bign tmp=1;

        for(int j=1;j<=n;j++) tmp=tmp*f[i-1];

        f[i]=f[i]+tmp+1;

    }

    bign ans=f[d]-f[d-1];

    cout<<ans;

}