這題本質上是 DP
為了加速所以寫了大數的快速冪
dp [i]=dp [i-1]^n+1 感謝網路博客 http://www.cnblogs.com/staginner/archive/2011/12/17/2290920.html
每次都有一個頭然後 n 個連出去的節點
所以每個節點都可以選深度 0~i-1 所有的嚴格 n 元數
可以選的數字就是 dp [i-1] 那麼就是 dp [i-1]^n 可以使用的排列組合
但是別忘記還有一個節點的情況 所以要 + 1
//
// GGGGGGGGGGGGG CCCCCCCCCCCCC AAA
// GGG::::::::::::G CCC::::::::::::C A:::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::::A
// G:::::G GGGGGG C:::::C CCCCCC A:::::::::A
//G:::::G C:::::C A:::::A:::::A
//G:::::G C:::::C A:::::A A:::::A
//G:::::G GGGGGGGGGGC:::::C A:::::A A:::::A
//G:::::G G::::::::GC:::::C A:::::A A:::::A
//G:::::G GGGGG::::GC:::::C A:::::AAAAAAAAA:::::A
//G:::::G G::::GC:::::C A:::::::::::::::::::::A
// G:::::G G::::G C:::::C CCCCCC A:::::AAAAAAAAAAAAA:::::A
// G:::::GGGGGGGG::::G C:::::CCCCCCCC::::C A:::::A A:::::A
// GG:::::::::::::::G CC:::::::::::::::C A:::::A A:::::A
// GGG::::::GGG:::G CCC::::::::::::C A:::::A A:::::A
// GGGGGG GGGG CCCCCCCCCCCCCAAAAAAA AAAAAAA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) decltype(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define gettime() end\_time=clock();printf("now running time is %.7f\\n",(float)(end\_time - start\_time)/CLOCKS\_PER\_SEC);
#else
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...)
#define gettime()
#endif
typedef unsigned int uint;
typedef long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 30005
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-6
#define X first
#define Y second
clock\_t start\_time=clock(), end\_time;
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
struct bignum{
Int n\[63\];
static const int maxn=60;
static const int base\=10000;
bignum(){
Set(n,0);
}
bignum(Int x){
Set(n,0);
n\[0\]=x;
}
bignum operator+(int a)const{
bignum ans;
Int c=0;
for(int i=0;i<maxn;i++){
ans.n\[i\]=n\[i\]+a+c;
c=ans.n\[i\]/base;
ans.n\[i\]%=base;
a=0;
}
return ans;
}
bignum operator\*(bignum &d)const{
bignum ans;
for(int i=0;i<maxn;i++){
if(n\[i\]){
for(int j=0;j+i<maxn;j++){
ans.n\[i+j\]+=n\[i\]\*d.n\[j\];
Int c=ans.n\[i+j\]/base;
ans.n\[i+j\]%=base;
for(int k=i+j+1;k<maxn;k++){
Int tmp=ans.n\[k\]+c;
c=tmp/base;
ans.n\[k\]=tmp%base;
}
}
}
}
return ans;
}
bignum operator\-(bignum &d)const{
Int c=0;
bignum ans;
for(int i=0;i<maxn;i++){
ans.n\[i\]=n\[i\]-d.n\[i\]-c;
if(ans.n\[i\]<0){
c=1;
ans.n\[i\]+=base;
}else c=0;
}
return ans;
}
void print(){
int i=maxn-1;
for(;i>0;i--)if(n\[i\])break;
printf("%lld",n\[i--\]);
for(;i>=0;i--)printf("%04lld",n\[i\]);
Pln();
}
};
bignum bpow(bignum a,int n){
bignum ans(1);
while(n){
if(n&1){
ans=ans\*a;
}
a=a\*a;
n>>=1;
}
return ans;
}
int n,d;
void solve(){
// bignum d(2);
// d.print();
bignum dp\[20\];
dp\[0\]=dp\[0\]+1;
for(int i=1;i<=d;i++){
dp\[i\]=bpow(dp\[i-1\],n);
dp\[i\]=dp\[i\]+1;
// dp\[i\].print();
}
printf("%d %d ",n,d);
if(d==0)dp\[0\].print();
else (dp\[d\]-dp\[d-1\]).print();
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%d%d",&n,&d)&&n+d){
solve();
}
}