bzoj1089: [SCOI2003]严格n元树

题解：http://hi.baidu.com/gzh19950711/item/c3ac0f0b0ad3b0c190571881

超级详细，我就不说了

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <stack>
#include <deque>
#include <queue>
#include <map>
#include <vector>
#include <set>
typedef long long LL;
typedef double DB;
typedef unsigned US;
#define For(i, s, t) for(int i = (s); i <= (t); i++)
#define Ford(i, s, t) for(int i = (s); i >= (t); i--)
#define Rep(i, n) for(int i = 0; i < (n); i++)
#define Repn(i, n) for(int i = ((n) - 1); i >= 0; i--)
#define rep(i, s, t) for(int i = (s); i < (t); i++)
#define repn(i, s, t) for(int i = ((s) - 1); i >= (t); i--)
#define FU(t, c) for(__typeof((c).begin()) t = (c).begin(); t != (c).end(); t++)
#define FD(t, c) for(__typeof((c).rbegin()) t = (c).rbegin(); t != (c).rend(); t--)
#define pi (3.1415926535)
#define mk make_pair
#define sqr(x) ((x) * (x))
#define ft first
#define sd second
#define abs(x) ((x) > 0 ? (x) : (-(x)))
#define max(x, y) ((x) > (y) ? (x) : (y))
#define min(x, y) ((x) < (y) ? (x) : (y))
#define clr(c, t) (memset((c), (t), sizeof((c))))
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define Sma_let(x) (((x) >= 'a') && ((x) <= 'z'))
#define Big_let(x) (((x) >= 'A') && ((x) <= 'Z'))
#define letter(x) ((Sma_let((x))) || (Big_let((x))))
#define MIT 2147483647
#define INF 1000000000
#define MLL 1000000000000000000LL
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
using namespace std; // yzx'Rp ++
inline void SETIO(string name) {
	string IN = name + ".in", OUT = name + ".out";
	freopen(IN.c_str(), "r", stdin), freopen(OUT.c_str(), "w", stdout);
}

const int B = 100000000, M = 2100, N = 60; 
struct BigNum {
	int a[M], len;
	BigNum() { len=0, clr(a, 0); }
	inline int operator [](int v) const { return a[v]; }
	inline int &operator [](int v) { return a[v]; }
	inline void operator =(int v) {
		len = 0;
		for (; v; v /= B) a[++len] = v % B;
	}
	inline void operator =(const BigNum &v) { len = v.len, memcpy(a, v.a, sizeof(v.a)); }
	inline BigNum(int v) : len(0) {
		while(v) a[++len] = v % B, v /= B;
	}
	inline BigNum(const BigNum &v) : len(v.len) { memcpy(a, v.a, sizeof(v.a)); }
} Dp[N][N];
int n, d;

inline void Input() {
	scanf("%d%d", &n, &d);
}

inline BigNum operator +(const BigNum &a, const BigNum &b) {
	BigNum r;
	int i = 0, c = 0;
	for (i = 1; i <= max(a.len, b.len) || c > 0; i++) {
		c += a[i] + b[i];
		r[i] = c % B, c /= B;
	}
	r.len = i - 1;
	return r;
}

inline BigNum operator -(const BigNum &a, const BigNum &b) {
	int c = 0;
	BigNum r;
	r.len = a.len;
	For(i, 1, r.len) {
	   r[i] = a[i] - c - b[i];
	   if(r[i] < 0) r[i] += B, c = 1;
	   else c = 0;
	}
	while(r.len > 0 && r[r.len] == 0) r.len--;
	return r;
}

inline BigNum operator *(const BigNum &a, const BigNum &b) {
	BigNum r;
	LL c = 0;
	For(i, 1, a.len)
		For(j, 1, b.len) {
	   		c = (LL) a[i] * (LL) b[j] + (LL) r[i + j - 1];
			r[i + j - 1] = c % B, r[i + j] += c / B;
		}
	r.len = a.len + b.len;
	while (r.len > 0 && r[r.len]==0) r.len--;
	return r;
}

inline BigNum operator ^(const BigNum &a, int b) {
	if(b == 1) return a;
	if(b == 0) return 1;
	BigNum Ans = a ^ (b >> 1);
	if(b % 2 == 0) return sqr(Ans);
	else return sqr(Ans) * a; 
}

ostream &operator <<(ostream &out, const BigNum &v) {
	out << v[v.len];
	Ford(i, v.len - 1, 1)
	   for(int j = B / 10; j; j /= 10) out << (v[i] / j) % 10;
	return out;
}

inline void Solve() {
	BigNum sum = 1;
	Dp[n][0] = sum;
	For(i, 1, d) Dp[n][i] = (sum ^ n) - sum + 1, sum = sum + Dp[n][i];
	cout << Dp[n][d] << endl;
}
int main() {
	#ifndef ONLINE_JUDGE
	SETIO("1089");
	#endif
	Input();
	Solve();
	return 0;
}