POJ

来源：互联网发布：windows rt升级win10 编辑：程序博客网时间：2024/05/21 06:34

题目：

求第n棵二叉树

思路：用卡特兰数确定一共有几个节点，然后递归输出

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fLL h[30];void Init(){    h[0]=h[1]=1;    for(int i=2;i<=25;i++)        h[i]=h[i-1]*(4*i-2)/(i+1);}void dfs(int k,int pos){    if(k==1){        printf("X");        return;    }    if(pos<=h[k-1]){//只有右子树        printf("X");        printf("(");        dfs(k-1,pos);        printf(")");        return;    }    else if(pos>h[k]-h[k-1]){//只有左子树        printf("(");        dfs(k-1,pos-(h[k]-h[k-1]));        printf(")");        printf("X");    }    else{        int t=k-1,m;        for(int i=t;i>=0;i--){            if(pos>h[t-i]*h[i])                pos-=h[t-i]*h[i];            else{                m=i;                break;            }        }        printf("(");        dfs(t-m,pos/h[m]+(pos%h[m]!=0));        printf(")");        printf("X");        printf("(");        dfs(m,(pos%h[m]==0)?h[m]:(pos%h[m]));        printf(")");    }}int main(){    int m,n;    Init();    while(~scanf("%d",&n)){        if(!n)            break;        for(int i=1;;i++){            if(n>h[i]){                n-=h[i];            }            else{                m=i;                break;            }        }        dfs(m,n);        printf("\n");    }    return 0;}

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