题目描述
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，如下图：
给出n, d，编程数出深度为d的n元树数目。
输入输出格式
输入格式：
仅包含两个整数n, d(0<n<=32, 0<=d<=16)。输入数据保证你不需要考虑某一层多于1024个节点的树（即nd<=1024）。提示：答案保证不超过200位十进制数。
输出格式：
仅包含一个数，即深度为d的n元树的数目。
输入输出样例
输入样例#1：
2 2
输出样例#1：
3
输入样例#2：
2 3
输出样例#2：
21
输入样例#3：
3 5
输出样例#3：
58871587162270592645034001
f[i] 为深度不小于i的树的总个数
先算出较浅深度的树的个数，然后每次^n+1(补一个根)
最后要深度为d的只需f[d]-f[d-1]即为答案
code:
//By Menteur_Hxy
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <vector>
#include <queue>
#include <set>
#include <ctime>
#define M(a,b) memset(a,(b),sizeof(a))
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
#define LL long long
using namespace std;
inline LL rd() {
LL x=0,fla=1; char c=' ';
while(c>'9'|| c<'0') {if(c=='-') fla=-fla; c=getchar();}
while(c<='9' && c>='0') x=x*10+c-'0',c=getchar();
return x*fla;
}
inline void out(LL x){
int a[25],wei=0;
if(x<0) putchar('-'),x=-x;
for(;x;x/=10) a[++wei]=x%10;
if(wei==0){ puts("0"); return;}
for(int j=wei;j>=1;--j) putchar('0'+a[j]);
putchar('
');
}
const int maxn=10010;
const int INF=0x3f3f3f3f;
int n,d;
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[40];
int main() {
n=rd();d=rd();
if(d==1&&n==1) return cout<<0,0;
if(!d) return cout<<1,0;
f[1]=1;
F(i,1,d) {
bign tp=1;
F(j,1,n) tp=tp*f[i-1];
f[i]=f[i]+tp+1;
}
return cout<<f[d]-f[d-1],0;
}