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Life, it seems, will fade away

drifting further every day

日志
2009/8/28

A better one

int hourMinuteHandInRight = 0;

int hourMinuteHandInLeft = 0;

int interval = 60;

int rightHand = 180 - interval;

int leftHand = 540 - interval;

while (hourMinuteHandInRight < 12 && hourMinuteHandInLeft < 12)

{

rightHand += interval;

if (rightHand == 660)

{

rightHand = 0;

hourMinuteHandInRight++;

}

Console.WriteLine("{0}:{1}", hourMinuteHandInRight, (int)(rightHand / 11));

hourMinuteHandInRight++;

leftHand += interval;

if (leftHand >= 660)

{

hourMinuteHandInLeft++;

leftHand = 0;

}

Console.WriteLine("{0}:{1}", hourMinuteHandInLeft , (int)(leftHand / 11));

hourMinuteHandInLeft++;

}
13:02:44 | 添加评论 | 固定链接 | 写入日志
2009/8/27

Write a function to give out all the time (hour:minute) that the angle between minute hand and hour hand on a clock is 90 degree.

static void Main(string[] args)

{

for (int h = 0; h < 12; h++)

{

double m = 1.0;

double angle = 1.0

if (h < 9)

{

m = 1.0*(60 * h + 180) / 11;

angle = (m - h * 5 - m / 12) * 6;

Console.WriteLine("{0}:{1} {2}", h, (int)m, angle);

}

if (h > 9)

{

m = 1.0 * (60 * h - 540) / 11;

angle = (h * 5 + m / 12 - m) * 6;

Console.WriteLine("{0}:{1} {2}", h, (int)m, angle);

}

if (h < 3)

{

m = 1.0 * (60 * h + 540) / 11;

angle = (m - h * 5 - m / 12) * 6;

Console.WriteLine("{0}:{1} {2}", h, (int)m, angle);

}

if (h > 3)

{

m = 1.0 * (60 * h - 180) / 11;

angle = (h * 5 + m / 12 - m) * 6;

Console.WriteLine("{0}:{1} {2}", h, (int)m, angle);

}

}

}

14:13:09 | 添加评论 | 固定链接 | 写入日志 | 计算机与 Internet
2009/8/4

Traverse binary tree not recursively

       public static void PreOrder<T>(BinaryTreeNode<T> root, Action<BinaryTreeNode<T>> action)
        {
            if (root == null)
            {
                return;
            }
            Stack<BinaryTreeNode<T>> stack = new Stack<BinaryTreeNode<T>>();
            stack.Push(root);
            while (stack.Count != 0)
            {
                BinaryTreeNode<T> current = stack.Pop();
                if (current != null)
                {
                    action(current);
                }
                if (current.RightChild != null)
                {
                    stack.Push(current.RightChild);
                }
                if (current.LeftChild != null)
                {
                    stack.Push(current.LeftChild);
                }
            }
        }
 
public static void InOrder<T>(BinaryTreeNode<T> root, Action<BinaryTreeNode<T>> action)
        {
            if (root == null)
            {
                return;
            }
            Stack<BinaryTreeNode<T>> stack = new Stack<BinaryTreeNode<T>>();
            BinaryTreeNode<T> current = root;
            do
            {
                while (stack.Count != 0 && current == null)
                {
                    current = stack.Pop();
                    action(current);
                    current = stack.Pop();
                }
                if (current != null)
                {
                    stack.Push(current.RightChild);
                    stack.Push(current);
                    current = current.LeftChild;
                }
            } while (stack.Count != 0);
        }
 
        public static void LastOrder<T>(BinaryTreeNode<T> root, Action<BinaryTreeNode<T>> action)
        {
            if (root == null)
            {
                return;
            }
            Stack<BinaryTreeNode<T>> stack = new Stack<BinaryTreeNode<T>>();
            BinaryTreeNode<T> current = root;
            do
            {
                if (current.HasChildren)
                {
                    stack.Push(current);
                    if (current.LeftChild != null)
                    {
                        if (current.RightChild != null)
                        {
                            stack.Push(current.RightChild);
                        }
                        current = current.LeftChild;
                    }
                    else
                    {
                        current = current.RightChild;
                    }
                }
                else
                {
                    action(current);
                    while (stack.Count != 0 && current.IsChildrenOf(stack.Peek()))
                    {
                        current = stack.Pop();
                        action(current);
                    }
                    if (stack.Count != 0)
                    {
                        current = stack.Pop();
                    }
                }
            } while (stack.Count != 0);
        }
22:56:19 | 添加评论 | 固定链接 | 写入日志

Link binary tree nodes at the same level

Given modified binary tree node
public class TreeNode<T>
{
    public TreeNode<T> LeftChild;
    public TreeNode<T> RightChild;
    public TreeNode<T> Right;
}
        public static void LinkEachLevel<T>(TreeNode<T> root)
        {
            LinkedList<TreeNode<T>> rightMost = new LinkedList<TreeNode<T>>();
            LinkedListNode<TreeNode<T>> first = new LinkedListNode<TreeNode<T>>(null);
            rightMost.AddFirst(first);
            LinkEachLevel(root, rightMost.First);
        }
 
        public static void LinkEachLevel<T>(TreeNode<T> node, LinkedListNode<TreeNode<T>> rightNode)
        {
            if (rightNode.Value != null)
            {
                node.Right = rightNode.Value;
            }
            rightNode.Value = node;
            if (rightNode.Next == null)
            {
                rightNode.List.AddLast(new LinkedListNode<TreeNode<T>>(null));
            }
            if (node.RightChild != null)
            {
                LinkEachLevel((TreeNode<T>)node.RightChild, rightNode.Next);
            }
            if (node.LeftChild != null)
            {
                LinkEachLevel((TreeNode<T>)node.LeftChild, rightNode.Next);
            }
        }
 
 
22:53:19 | 添加评论 | 固定链接 | 写入日志 | Computers and Internet
2008/9/1

Reporting Services中如何对数据进行分页合计?

Report services中不直接支持分页合计,实现这个功能得加些vb code(report code).
  • 右键report区域(BIDS中最外层的黄色区域),选择report properties;打开code tab.
  • 加入如下代码

public shared Dim _rowCount As Int32=0
public shared Dim _sum as Int32=0

public Function AddToRowCount() as Int32
 _rowCount = _rowCount + 1
 return _rowCount
End Function

public Function AddToSum(ByVal quantity as Int32) as Int32
 _sum+=quantity
 return quantity
End Function

public function GetSum() as Int32
 Dim currentSum= _sum
 _sum=0
 return currentSum
end function

public Function GetCurrentCountAndReset() As Int32
  Dim currentRowCount = _rowCount
  _rowCount = 0
  Return currentRowCount
End Function

  • 实现row index.

添加一个新列,设置它的expression为=Code.AddToRowCount

  • 实现quantity sum.

设计Quantity列(显示数额的列)的expression为=Code.AddToSum(Fields!Quantity.Value)

在Footer加入textbox, 并设置表达式为=Code.GetSum

12:36:00 | 添加评论 | 固定链接 | 写入日志 | 计算机与 Internet