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(2011-05-04 12:27)

### 杂谈

Description

Strategies for compressing two-dimensional images are often based on finding regions with high similarity. In this problem, we explore a particular approach based on a hierarchical decomposition of the image. For simplicity, we consider only bitmapped images such as the following:
The image is encoded as a tree, with the root representing the entire image region. If a region is monochromatic, then the node for that region is a leaf storing the color of the region. Otherwise, the region is divided into four parts about its center, and the approach is applied recursively to each quadrant. For a non-leaf node, its four children represent the four quadrants ordered as upper-right, upper-left, lower-left, lower-right respectively. As an example, here is the tree encoding

(2011-04-29 21:09)

### it

Triangle Conjecture

Problem Description
One could construct a triangle with the digit 1 to 9 as the figure below:

The triangle is the one that the sums of every four numbers on its three edges are all equals to 23. Moreover, 23 is the biggest summation one can get from this kind of arraignment of digits. Your task is even tougher, given a positive integer n, you should use integer from 1 to 3*(n-1) to construct triangle with equal summation of digits on the three edges and the summation is the biggest among all the possible arraignments. For example, if n = 4, then you should choose number from 1 to 3*(4-1).
For convenience, the output format for a certain triangle is like the example for the fi

(2011-04-28 16:49)

### it

(2005广东省赛) 椭圆相交

Recently the astronomers have discovered a peculiar pair of planets, named A and B. As we know, a planet usually moves in an ellipse orbit, so do A and B. But their orbits are quite special: (1) Their orbits are in the

(2011-04-24 14:39)

### it

(2011-04-16 18:16)

### 除

import java.util.*;

import java.math.*;

(2011-04-11 16:00)

### it

#include <iostream>
using namespace std;

void Move( char A, char C )
{
cout << A << ' --> ' << C << endl;
}

void Hanoi( int n, char A, char B, char C )
{
if( n == 1 )
{
Move( A, C );
}
else
{
Hanoi( n-1, A, C, B );
Move( A, C );
Hanoi( n-1, B, A, C );
}
}

int _tmain(int argc, _TCHAR* argv[])
{
int disks;

cout << '请输入汉诺塔盘子的个数：' << endl;
cin >> disks;
Hanoi( disks, '1', '2', '3' );

return 0;
}

(2011-04-10 22:21)

### it

POJ 3233   Matrix Power Series

Description

Given a n × n matrix A and a positive integer k, find the sum S

(2011-03-25 21:26)

### it

import javax.swing.JOptionPane;

public class HanoiResolve {
public static int count = 0; //步骤计数器
public static void main(String[] args) {
String numberString = JOptionPane.showInputDialog('Enter the number of the disks you want me move.');
int number = Integer.parseInt(numberString);
Hanoi( number, 'A', 'B', 'C' );
}

public static void Hanoi( int n, char A, char B, char C )
{
if( n == 1 )
Move( A, C );
else
{
Hanoi( n-1, A, C, B );
Move( A, C );
Hanoi( n-1, B, A, C );
&

(2011-01-03 22:14)

### 杂谈

(2010-12-31 15:46)