• hdu1130 How Many Trees?(Catalan+高精)


    Problem Description

    A binary search tree is a binary tree with root k such that any node v reachable from its left has label (v) < label (k) and any node w reachable from its right has label (w) > label (k). It is a search structure which can find a node with label x in O(n log n) average time, where n is the size of the tree (number of vertices).

    Given a number n, can you tell how many different binary search trees may be constructed with a set of numbers of size n such that each element of the set will be associated to the label of exactly one node in a binary search tree?

    Input

    The input will contain a number 1 <= i <= 100 per line representing the number of elements of the set.

    Output

    You have to print a line in the output for each entry with the answer to the previous question.

    Sample Input
    1
    2
    3

    Sample Output
    1
    2
    5

    Source
    UVA

    分析:
    给出结点数,求形态不同的二叉树
    Catalan经典模型

    tip

    我要有逼格
    我要重载运算符(然而完全是多余)

    第一次交上去是PE
    我整个人都是mb的
    后来把输出改成了这个样子,就没有问题了

     for (int i=a[n].l;i>=1;i--)
         printf("%d",a[n].s[i]);
     puts("");

    习惯了忽略行末空格
    所以在做UVa的题的时候就比较难受

    //这里写代码片
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    
    using namespace std;
    
    struct node{
        int l,s[102];
        void clear()
        {
            memset(s,0,sizeof(s));
        }
    };
    node a[102];
    
    node operator *(const node &a,const int &x)
    {
        node b;
        b.clear();
        int ll=a.l;
        int d=0;
        for (int i=1;i<=ll;i++)
        {
            b.s[i]=a.s[i]*x+d;
            d=b.s[i]/10;
            b.s[i]%=10;
        }
    
        while (d)
        {
            b.s[++ll]=d;
            d=b.s[ll]/10;
            b.s[ll]%=10;        
        }
    
        b.l=ll;
        return b;
    }
    
    node operator /(const node &a,const int &x)
    {
        node b;
        b.clear();
        int d=0;
        for (int i=a.l;i>=1;i--)
        {
            int t=d*10+a.s[i];
            b.s[i]=t/x;
            d=t%x;
        }
    
        int ll=a.l;
        while (b.s[ll]==0) ll--;
    
        b.l=ll;
        return b;
    }
    
    void Catalan()
    {
        a[1].clear();
        a[1].l=1; a[1].s[1]=1;
        for (int i=2;i<=100;i++)
        {
            a[i]=a[i-1]*(4*i-2);
            a[i]=a[i]/(i+1);
        }
    }
    
    int main()
    {
        Catalan();
        int n;
        int cnt=0;
        while (scanf("%d",&n)!=EOF)
        {
            for (int i=a[n].l;i>=1;i--)
                printf("%d",a[n].s[i]);
            puts("");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wutongtong3117/p/7673055.html
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