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来自洛谷，原作者为

	GhostCai **

搬运于2025-08-24 21:58:40，当前版本为作者最后更新于2018-06-04 17:55:29，作者可能在搬运后再次修改，您可在原文处查看最新版

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以下是正文

f[i]表示深度不超过i的严格n元树的个数。
转移时考虑加一个根，由于是n元树，根下需要n个子树
f[i]=(f[i-1]^n)+1 答案即为对f[d]-f[d-1] 正体现了差分与前缀和的神奇作用

边界f[0]=1，且当d=0时特判答案为1

需要高精度

#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[17];
int n,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;
}