UVA 10516 Another Counting Problem

大意略。

思路：组合数学神马的，最难思考了。

我想这道题的结论可以记下来，即：满K叉树的前i层的不同的树的数量为：f[i-1]^K + 1

参考了这篇博客：解法

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cmath>
using namespace std;

const int MAXN = 1010;

int base, n;

struct bign
{
	int s[MAXN], len;
	bign () {memset(s, 0, sizeof(s)); len = 1;}
	bign (int num) { *this = num;}
	bign (const char *num) {*this = num;}
	bign operator = (int num)
	{
		char s[MAXN];
		sprintf(s, "%d", num);
		*this = s;
		return *this;
	}
	bign operator = (const char *num)
	{
		len = strlen(num);
		for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
		return *this;
	}
	bign operator + (const bign &b)
	{
		bign c;
		c.len = 0;
		for(int i = 0, g = 0; 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 % 10;
			g = x / 10;
		}
		return c;
	}
	void clean()
	{
		while(len > 1 && !s[len-1]) len--;
	}
	bign operator - (const bign &b)
	{
		bign c;
		c.len = 0;
		for(int i = 0, g = 0; i < len; i++)
		{
			int x = s[i] - g;
			if(i < b.len) x -= b.s[i];
			if(x >= 0) g = 0;
			else
			{
				g = 1;
				x += 10;
			}
			c.s[c.len++] = x;
		}
		c.clean();
		return c;
	}
	bign operator * (const bign &b)
	{
		bign c;
		c.len = len + b.len;
		for(int i = 0; i < len; i++)
		{
			for(int j = 0; j < b.len; j++)
			{
				c.s[i+j] += s[i] * b.s[j];
			}
		}
		for(int i = 0; i < c.len; i++)
		{
			c.s[i+1] += c.s[i] / 10;
			c.s[i] %= 10;
		}
		c.clean();
		return c;
	}
	bign operator *= (const bign &b)
	{
		*this = *this * b;
		return *this;
	}
	string str() const
	{
		string res = "";
		for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
		if(res == "") res = "0";
		return res;
	}
}f[2020];

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

void cal()
{
	f[0] = 1;
	for(int i = 1; i <= n; i++)
	{
		bign temp = 1;
		for(int j = 1; j <= base; j++)
		{
			temp *= f[i-1];
		}
		f[i] = temp+1;
	}
}

void solve()
{
	printf("%d %d ", base, n);
	cout<<f[n]-f[n-1]<<endl;
}

int main()
{
	while(~scanf("%d%d", &base, &n) && (base || n))
	{
		cal();
		solve();
	}
	return 0;
}