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。
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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。
仅包含两个整数n, d( 0 < n < = 32, 0 < = d < = 16)
仅包含一个数,即深度为d的n元树的数目。
#define mx 300 #include<cstdio> #include<iostream> using namespace std; struct gaojing{ int len; int a[mx+10]; }zero,one,f[20]; int n,d; inline void set0(gaojing &s)//¸ß¾«ÇåÁã { s.len=1; for (int i=1;i<=mx+5;i++)s.a[i]=0; } inline void inputn(gaojing &a)//¸ß¾«ÊäÈë { set0(a); char ch=getchar(); while (ch<‘0‘||ch>‘9‘)ch=getchar(); while (ch>=‘0‘&&ch<=‘9‘) { a.a[a.len++]=ch-‘0‘; ch=getchar(); } a.len--; int change[mx+15]; for (int i=1;i<=a.len;i++) change[i]=a.a[i]; for (int i=1;i<=a.len;i++) a.a[i]=change[a.len-i+1]; while (a.a[a.len]==0)a.len--; } inline void put(gaojing a)//¸ß¾«Êä³ö { for (int i=a.len;i>=1;i--)printf("%d",a.a[i]); printf("\n"); } inline bool operator < (const gaojing &a,const gaojing &b)//¸ß¾«< { if (a.len<b.len)return 1; if (a.len>b.len)return 0; for (int i=a.len;i>=1;i--) { if (a.a[i]<b.a[i])return 1; if (a.a[i]>b.a[i])return 0; } return 0; } inline bool operator == (const gaojing &a,const gaojing &b)//¸ß¾«> { if (a.len!=b.len)return 0; for (int i=a.len;i>=1;i--) { if (a.a[i]!=b.a[i])return 0; } return 1; } inline gaojing max(const gaojing &a,const gaojing &b)//¸ß¾«max { if (a<b)return b; else return a; } inline gaojing min(const gaojing &a,const gaojing &b)//¸ß¾«min { if (a<b)return a; else return b; } inline gaojing operator + (const gaojing &a,const gaojing &b)//¸ß¾«+ { gaojing c;set0(c); int maxlen=max(a.len,b.len); for (int i=1;i<=maxlen;i++) { c.a[i]=c.a[i]+a.a[i]+b.a[i]; if (c.a[i]>=10) { c.a[i+1]+=c.a[i]/10; c.a[i]%=10; } } c.len=maxlen+4; while (!c.a[c.len]&&c.len>1) c.len--; return c; } inline gaojing operator - (const gaojing &a,const gaojing &b)//¸ß¾«- { gaojing c;set0(c); gaojing d;d=a; for (int i=1;i<=b.len;i++) { c.a[i]=d.a[i]-b.a[i]; if (c.a[i]<0) { c.a[i]+=10; int now=i+1; while (!d.a[now]) { d.a[now]=9; now++; } d.a[now]--; } } for (int i=b.len+1;i<=d.len;i++)c.a[i]=d.a[i]; c.len=d.len; while (c.a[c.len]==0&&c.len>1)c.len--; return c; } inline gaojing operator * (const gaojing &a,const gaojing &b)//¸ß¾«* { gaojing c;set0(c); for(int i=1;i<=a.len;i++) for (int j=1;j<=b.len;j++) c.a[i+j-1]+=a.a[i]*b.a[j]; c.len=a.len+b.len+5; for (int i=1;i<=c.len;i++) { c.a[i+1]+=c.a[i]/10; c.a[i]%=10; } while (!c.a[c.len]&&c.len>1)c.len--; return c; } inline void div_by_2(gaojing &a) { for (int i=a.len;i>=1;i--) { if (a.a[i]&1 && i!=1)a.a[i-1]+=10; a.a[i]/=2; } while (!a.a[a.len]&&a.len>1)a.len--; } inline gaojing operator / (gaojing a,const gaojing &b)//¸ß¾«/ { gaojing l,r,ans; set0(l);l.len=1; set0(r);r=a; set0(ans);ans.len=1; while (l<r||l==r) { gaojing mid=l+r; div_by_2(mid); if(mid*b==a)return mid; if(mid*b<a){ans=mid;l=mid+one;} if(a<mid*b)r=mid-one; } return ans; } inline gaojing operator ^(const gaojing &a,int p)//¸ß¾«^ { gaojing ans=one,mult=a; while (p) { if (p&1)ans=ans*mult; mult=mult*mult; p>>=1; } return ans; } inline void chushihua()//³õʼ»¯£¬¶Ô0¡¢1¸ß¾«¶È³£Êý¸³Öµ { set0(zero); zero.len=1; set0(one);one.len=1;one.a[1]=1; } int main() { chushihua(); scanf("%d%d",&n,&d); f[0]=one; for (int i=1;i<=d;i++) { f[i]=f[i-1]^n; f[i]=f[i]+one; } put(f[d]-f[d-1]); }
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原文地址:http://www.cnblogs.com/zhber/p/4098407.html