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左言 — Sun, 12 Dec 2010 16:19:00 GMT

#include <iostream>
#include <cmath>
using namespace std;

/*
long long max(long long n)
{
    long long i=2;
    while(n != 1)
    {
        while (n % i == 0)
            n /= i;
        i++;
    }
    return i-1;
}
*/

// return the max prime factor of n
long long max(long long n)
{
    long long i=2;
    long long m = (long long)sqrt(n);
    while(i <= m)
    {
        if (n % i == 0)
        {
            n /= i;
            m = (long long)sqrt(n);
        }
        else {
            i++;
        }
    }
    return n;
}

int main()
{
    cout << max(13195) << endl;
    cout << max(600851475143LL) << endl;
    return 0;
}

左言 2010-12-13 00:19 发表评论

二叉树非递归遍历

左言 — Fri, 22 Oct 2010 06:23:00 GMT

#include <iostream>
#include <stack>
#include <map>
using namespace std;

struct Node {
    int v;
    Node *lchild;
    Node *rchild;
};

struct StackNode {
    Node *ptr;
    int flag; // 1 means the right sub-tree has been travelled
};

void visit(Node *p)
{
    cout << p->v << ' ';
}

void preorder_travel(Node *root)
{
    if(root == NULL)
        return;
    stack<Node*> s;
    Node *p = root;
    while (p || !s.empty())
    {
        while (p)
        {
            s.push(p);
            visit(p);
            p = p->lchild;
        }
        if (!s.empty())
        {

            p = s.top();
            s.pop();
            p = p->rchild;
        }
    }
}

void inorder_travel(Node *root)
{
    if(root == NULL)
        return;
    stack<Node*> s;
    Node *p = root;
    while (p || !s.empty())
    {
        while (p)
        {
            s.push(p);
            p = p->lchild;
        }
        if (!s.empty())
        {
            p = s.top();
            s.pop();
            visit(p);
            p = p->rchild;
        }
    }
}

void postorder_travel(Node *root)
{
    if(root == NULL)
        return;
    stack<StackNode> s;
    StackNode x;

    Node *p = root;
    while (p || !s.empty())
    {
        while (p)
        {
            x.ptr = p;
            x.flag = 0;
            s.push(x);
            p = p->lchild;
        }
        while (!s.empty() && s.top().flag)
        {
            visit(s.top().ptr);
            s.pop();
        }
        if (!s.empty())
        {
            s.top().flag = 1;
            p = s.top().ptr->rchild;
        }
    }
}

int main()
{
    int i, j;
    map<int, Node*> m;

    int a[][3] = {
        {1,2,3},
        {2,4,5},
        {3,6,7},
        {4,-1,-1},
        {5,-1,-1},
        {6,8,-1},
        {7,-1,9},
        {8,-1,-1},
        {9,-1,-1},
        -1
    };

    for (i=0; a[i][0]!=-1; i++) {
        Node *n = new Node;
        n->v = a[i][0];
        m[n->v] = n;
    }
    for (i=0; a[i][0]!=-1; i++) {
        int s = a[i][0];
        int l = a[i][1];
        int r = a[i][2];
        Node *n = m[s];
        n->lchild = (l==-1 ? NULL : m[l]);
        n->rchild = (r==-1 ? NULL : m[r]);
    }

    Node *root = m[a[0][0]];

    preorder_travel(root);
    cout << endl;
    inorder_travel(root);
    cout << endl;
    postorder_travel(root);
    cout << endl;

    return 0;
}

左言 2010-10-22 14:23 发表评论

左言 — Fri, 22 Oct 2010 06:16:00 GMT

使用非递归后箋遍历�Q?/div>

#include<iostream>
#include<stack>
#include<map>
using namespace std;

struct Node
{
    int v;
    Node *lchild;
    Node *rchild;
};

struct StackNode
{
    Node *ptr;
    int flag;
};

int maxDepth(Node *root)
{
    stack<StackNode> s;
    StackNode x;
    Node *p = root;
    int max = 0;
    while (p || !s.empty())
    {
        while (p)
        {
            x.ptr = p;
            x.flag = 0;
            s.push(x);
            if (s.size() > max)
                max = s.size();
            p = p->lchild;
        }
        while (!s.empty() && s.top().flag)
        {
            s.pop();

        }
        if (!s.empty())
        {
            s.top().flag = 1;
            p = s.top().ptr->rchild;
        }
    }
    return max;
}

int main()
{
    int i, j;
    map<int, Node*> m;

    int a[][3] = {
        {1,2,3},
        {2,4,5},
        {3,6,7},
        {4,-1,-1},
        {5,-1,-1},
        {6,8,-1},
        {7,-1,9},
        {8,-1,-1},
        {9,-1,10},
        {10,-1,-1},
        -1
    };

    for (i=0; a[i][0]!=-1; i++) {
        Node *n = new Node;
        n->v = a[i][0];
        m[n->v] = n;
    }
    for (i=0; a[i][0]!=-1; i++) {
        int s = a[i][0];
        int l = a[i][1];

        int r = a[i][2];
        Node *n = m[s];
        n->lchild = (l==-1 ? NULL : m[l]);
        n->rchild = (r==-1 ? NULL : m[r]);
    }

    Node *root = m[a[0][0]];

    cout << maxDepth(root) << endl;

    return 0;
}

左言 2010-10-22 14:16 发表评论

左言 — Thu, 10 Jun 2010 04:16:00 GMT

摘要: 实现�q�个函数�Q?
char* strrep(const char* src, const char* from, const char* to);

��src中出现的所有from替换成to 阅读全文

左言 2010-06-10 12:16 发表评论

2009�q�全球网站Top50

左言 — Thu, 27 Aug 2009 03:27:00 GMT

摘要: ��国《时代》周刊评��Z��2009�q��?0个最佳网站”，都是哪些�|�站上榜�Q�它们具有哪些特点？阅读全文

左言 2009-08-27 11:27 发表评论

左言 — Fri, 14 Aug 2009 06:11:00 GMT

摘要: 最�q�一个朋友在用计��机模拟人来玩文曲星上的“猜数字”游戏，也邀我一赯��论。我以前没有玩过�Q�不�q�发现这个还是蛮有趣的，于是也想写一个能猜数据的��程序。通过几日思考，�l�于实现了一个稍微有点聪明的�Q�可以保证在7�ơ之内猜�Ҏ(gu��)��字�?nbsp; 阅读全文

左言 2009-08-14 14:11 发表评论

Ruby的一点小心得

左言 — Mon, 10 Aug 2009 16:09:00 GMT

摘要: 最�q�因��Z��用Ruby写一个多�U�程爬虫�Q�所以积累了一点小心得�?nbsp; 阅读全文

左言 2009-08-11 00:09 发表评论

左言 — Sat, 08 Aug 2009 02:47:00 GMT

摘要: 今天分析数据的时候需要检查一个图是否有向无环图，于是用脚本写了个��的��查工兗��基本原理：如果能够完全拓扑排序�Q�则是无环图�?
阅读全文

左言 2009-08-08 10:47 发表评论