# 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;



## 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;
}
posted @ 2018-06-05 14:25  Monster_Qi  阅读(108)  评论(0编辑  收藏  举报