其他

K-th NumberTime Limit: 20000MS Memory Limit: 65536KTotal Submissions: 20110 Accepted: 5251Case Time Limit: 2000MSDescriptionYou are working for Macrohard company in data structures department. After failing your previous task about key insertion you

A計劃Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2962    Accepted Submission(s): 634Problem

DivisibilityTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 524    Accepted Submission(s): 171Problem DescriptionAs we know,the fzu AekdyCoin is famous of math,especially in the field of number

題目描述：已經順序給出多邊形的各個頂點，求重心 #include <iostream>#include <algorithm>#include <cmath>using namespace std;struct point{int x, y;};point p[1000001];double cal(int i, int j){double res;res = p[i].x*p[j].y - p[j].x*p[i].y;return res;}int

CupTime Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 486    Accepted Submission(s): 150Problem DescriptionThe WHU ACM Team has a big cup, with which every member drinks water. Now, we know the

 GPATime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 317    Accepted Submission(s): 194Problem DescriptionEach course grade is one of the following five letters: A, B, C, D, and F. (Note that

如下代碼可以創造一個非模態對話方塊void CllView::OnDialog() { CTestDlg *pDlg = new CTestDlg;  pDlg->Create(IDD_DIALOG1,this);  pDlg->ShowWindow(SW_SHOW);}此時建立的對話方塊處於啟用狀態，是時候我們需要父視窗為啟用狀態而且不出現標題框的閃爍，我們只需要做到以下幾處改動：1，聲明p_Dlg為成員變數CTestDlg *pDlg，並為它申請空間pDlg = new

More is betterTime Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)Total Submission(s): 1576    Accepted Submission(s): 590Problem DescriptionMr Wang wants some boys to help him with a project. Because the project is

在VC6中加入對話方塊，並放入編輯框，運行後，在編輯框中輸入一些字母並按斷行符號，程式自動結束；在用Listctrl控制項時，對某一項進行編輯後習慣按斷行符號鍵進行確定，但是斷行符號鍵一按下，程式就自動結束了。對於這種問題，我們可以用如下簡單的辦法來解決：方法：重載CDialog::PreTranslateMessage(MSG* pMsg)；[程式實現]BOOL CMyDlg::PreTranslateMessage(MSG* pMsg) { if(pMsg->message ==

Can you find it?Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 1836    Accepted Submission(s): 400Problem DescriptionGive you three sequences of numbers A, B, C, then we give you a number X.

Mayor's postersTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14306 Accepted: 3997DescriptionThe citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at

 TilingTime Limit:1sMemory limit:32MAccepted Submit:197Total Submit:496In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles? Here is a sample tiling of a 2x17 rectangle.  Input is a sequence of lines, each line containing an integer

  Name PK Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 148    Accepted Submission(s): 75Problem DescriptionName PK is a funny game on the Internet. The game will calculate character's

GCDTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1323    Accepted Submission(s): 449Problem DescriptionGiven 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) =

【轉載MatRush@ZJUT百度部落格的】   矩陣運算是屬於線性代數裡的一個重要內容，上學期學完後只覺得矩陣能解線性方程，不過高中的時候聽說過矩陣能最佳化常係數遞推以及將座標上的點作線性變換，於是找了些資料研究了一下，並把許多經典題以及HDU shǎ崽大牛總結的矩陣乘法的題目[1]、[2]和開設的矩陣乘法DIY Contest給做完了，感覺收穫頗豐。    一個矩陣就是一個二維數組，為了方便聲明多個矩陣，我們一般會將矩陣封裝一個類或定義一個矩陣的結構體，我採用的是後者：   

1297. PalindromeTime Limit: 1.0 secondMemory Limit: 16 MBThe “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S. Robots»

原文：http://blog.csdn.net/hechunjiao2011/article/details/8262217MFC對話方塊應用程式本來是可以使用滑鼠來移動表單的，但是如果去掉對話方塊的標題列屬性就不能移動表單了，所以需要自己來實現滑鼠在客戶區點擊能移動表單的功能。    實現的原理就是：通過添加兩個訊息處理函數WM_NCHITTEST和WM_LBUTTONDOWN，前者的作用是可以將按一下滑鼠的客戶區當做標題列區，後者的作用是獲得按一下滑鼠的座標。   

bool isSqr(int n){    int a = int(sqrt(n) + 0.5);     return a * a == n;}bool isSqr(int n) {     int a = (int)(sqrt(n) + 0.0001);     return a * a == n; } bool isSqr(int n) {     int a = (int)sqrt(n * 1.0);     if(a*a == n || (a+1)*(a+1) == n || (a-1

最大報銷額Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6227    Accepted Submission(s): 1454Problem

Problem of PrecisionTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 240    Accepted Submission(s): 117Problem Description  InputThe first line of input gives the number of cases, T. T test