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DirectX Tutroial 1 First Window

2016-10-08

DirectX

CPP DirectX Graphics

Steps

This tutorials are based on the DirectX Tutorial, show how to set up IDE for using DirectX 9.

Steps

The basic Direct3D Program, includes four basic steps:

Create the global variable and function prototypes

// include the basic windows header files and the Direct3D header file
#include <windows.h>
#include <windowsx.h>
#include <d3d9.h>

// include the Direct3D Library file
#pragma comment (lib, "d3d9.lib")

// global declarations
LPDIRECT3D9 d3d;    // the pointer to our Direct3D interface
LPDIRECT3DDEVICE9 d3ddev;    // the pointer to the device class

// function prototypes
void initD3D(HWND hWnd);    // sets up and initializes Direct3D
void render_frame(void);    // renders a single frame
void cleanD3D(void);    // closes Direct3D and releases memory

// the WindowProc function prototype
LRESULT CALLBACK WindowProc(HWND hWnd, UINT message, WPARAM wParam, LPARAM lParam);

Create a function to initialize Direct3D and create the Direct3D Device

// this function initializes and prepares Direct3D for use
void initD3D(HWND hWnd)
{
    d3d = Direct3DCreate9(D3D_SDK_VERSION);    // create the Direct3D interface

    D3DPRESENT_PARAMETERS d3dpp;    // create a struct to hold various device information

    ZeroMemory(&d3dpp, sizeof(d3dpp));    // clear out the struct for use
    d3dpp.Windowed = TRUE;    // program windowed, not fullscreen
    d3dpp.SwapEffect = D3DSWAPEFFECT_DISCARD;    // discard old frames
    d3dpp.hDeviceWindow = hWnd;    // set the window to be used by Direct3D

    // create a device class using this information and information from the d3dpp stuct
    d3d->CreateDevice(D3DADAPTER_DEFAULT,
          D3DDEVTYPE_HAL,
          hWnd,
          D3DCREATE_SOFTWARE_VERTEXPROCESSING,
          &d3dpp,
          &d3ddev);
}

Create a function to render a frame

// this is the function used to render a single frame
void render_frame(void)
{
    // clear the window to a deep blue
    d3ddev->Clear(0, NULL, D3DCLEAR_TARGET, D3DCOLOR_XRGB(0, 40, 100), 1.0f, 0);

    d3ddev->BeginScene();    // begins the 3D scene

    // do 3D rendering on the back buffer here

    d3ddev->EndScene();    // ends the 3D scene

    d3ddev->Present(NULL, NULL, NULL, NULL);    // displays the created frame
}

Create a function to close Direct3D

// this is the function that cleans up Direct3D and COM
void cleanD3D(void)
{
    d3ddev->Release();    // close and release the 3D device
    d3d->Release();    // close and release Direct3D
}

You can get the complete codes from Github

If you run this program you should get something like this:

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hihoCoder 1082 然而沼跃鱼早就看穿了一切

2016-10-07

hihoCoder

hihoCoder String

Description
Input and Output
Codes

描述#1082 : 然而沼跃鱼早就看穿了一切

Description

fjxmlhx每天都在被沼跃鱼刷屏，因此他急切的找到了你希望你写一个程序屏蔽所有句子中的沼跃鱼(“marshtomp”，不区分大小写)。为了使句子不缺少成分，统一换成 “fjxmlhx” 。

Input and Output

输入

输入包括多行。

每行是一个字符串，长度不超过200。

一行的末尾与下一行的开头没有关系。

输出

输出包含多行，为输入按照描述中变换的结果。

样例输入

The Marshtomp has seen it all before.

marshTomp is beaten by fjxmlhx!

AmarshtompB

样例输出

The fjxmlhx has seen it all before.

fjxmlhx is beaten by fjxmlhx!

AfjxmlhxB

Codes

#include <iostream>
#include <string>
#include <algorithm> 
#include <cctype>

using namespace std;

int main() {
    string str, temp;
    const string s1 = "marshtomp";
    const string s2 = "fjxmlhx";
    
    string::size_type a = s1.size();
    string::size_type b = s2.size();
    while (getline(cin, temp)) {
        str = temp;
        transform(str.begin(),str.end(),str.begin(),::tolower);
        
        string::size_type pos = 0;
        while ((pos=str.find(s1,pos))!=string::npos) {
            str.replace(pos, a, s2);
            temp.replace(pos,a,s2);
            pos+=b;
        }
        cout << temp << endl;
    }
    
    return 0;
}

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hihoCoder 1186 Coordinates

2016-10-06

hihoCoder

hihoCoder
描述#1186 : Coordinates

Description

Give you two integers P and Q. Let all divisors of P be X-coordinates. Let all divisors of Q be Y-coordinates.

For example, when P=6 and Q=2, we can get the coordinates (1,1) (1,2) (2,1) (2,2) (3,1) (3,2) (6,1) (6,2).

You should print all possible coordinates in the order which is first sorted by X-coordinate when coincides, sorted by Y-coordinate.

Input and Output

输入

One line with two integers P and Q(1 <= P, Q <= 10000).

输出

The output may contains several lines , each line with two integers Xi and Yi, denoting the coordinates.
```
样例输入
	6 2

样例输出
	1 1
	1 2
	2 1
	2 2
	3 1
	3 2
	6 1
	6 2
```
Codes
```
#include <iostream>
#include <vector>

using namespace std;

int main () {
	int p, q;
	
	while (cin >> p >> q) {
		vector<int> arr1, arr2;

	for (int i=1; i<=p; i++)
		if (p % i==0)
			arr1.push_back(i);
	for (int i=1; i<=q; i++)
		if (q % i==0)
			arr2.push_back(i);

	for (int m=0; m<arr1.size(); m++)
		for (int n=0; n<arr2.size(); n++)
			cout << arr1[m] << " " << arr2[n] << endl;
	}
	
	return 0;
}
```
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2017-08-04

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hihoCoder 1148 2月29日

2016-10-04

hihoCoder

Description
Input and Output
Code

描述#1148 : 2月29日

Description

给定两个日期，计算这两个日期之间有多少个2月29日（包括起始日期）。

只有闰年有2月29日，满足以下一个条件的年份为闰年：

年份能被4整除但不能被100整除
年份能被400整除

Input and Output

输入

第一行为一个整数T，表示数据组数。

之后每组数据包含两行。每一行格式为”month day, year”，表示一个日期。month为{“January”, “February”, “March”, “April”, “May”, “June”, “July”, “August”, “September”, “October”, “November” , “December”}中的一个字符串。day与year为两个数字。

数据保证给定的日期合法且第一个日期早于或等于第二个日期。

输出

对于每组数据输出一行，形如”Case #X: Y”。X为数据组数，从1开始，Y为答案。

数据范围

1 ≤ T ≤ 550

小数据：

2000 ≤ year ≤ 3000

大数据：

2000 ≤ year ≤ 2×109

样例输入

	4
	January 12, 2012
	March 19, 2012
	August 12, 2899
	August 12, 2901
	August 12, 2000
	August 12, 2005
	February 29, 2004
	February 29, 2012

样例输出

	Case #1: 1
	Case #2: 0
	Case #3: 1
	Case #4: 3

Code

#include <iostream>
#include <cstring>

using namespace std;

int main() {
    int n;
    cin >> n;
    
    int startDay, endDay, startYear, endYear;
    char startMonth[20], endMonth[20];
    int i = 1;
    while (n--) {
        int count = 0;
        scanf("%s %d, %d", startMonth, &startDay, &startYear);
        scanf("%s %d, %d", endMonth, &endDay, &endYear);
        if (strcmp(startMonth, "February")!= 0 && strcmp(startMonth, "January")!= 0)
            startYear++;
        if ((strcmp(endMonth, "February")== 0 && endDay<29) || strcmp(endMonth, "January")== 0)
            endYear--;
        count = (endYear / 4 - endYear / 100 + endYear / 400) - (startYear / 4 - startYear / 100 + startYear / 400);
        if ((startYear%4==0 && startYear%100 !=0) || (startYear%400 ==0))
            count++;
        cout << "Case #" << i << ": " << count << endl;
        i++;
    }
    
    return 0;
}

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hihoCoder 1152 Lucky Substrings

2016-10-03

hihoCoder

Description
Input and Output
Code

描述#1152 : Lucky Substrings

Description

A string s is LUCKY if and only if the number of different characters in s is a fibonacci number. Given a string consisting of only lower case letters, output all its lucky non-empty substrings in lexicographical order. Same substrings should be printed once.

Input and Output

输入

A string consisting no more than 100 lower case letters.

输出

Output the lucky substrings in lexicographical order, one per line. Same substrings should be printed once.

样例输入
	aabcd

样例输出
	a
	aa
	aab
	aabc
	ab
	abc
	b
	bc
	bcd
	c
	cd
	d

Code

#include <iostream>
#include <string>
#include <set>
#include <unordered_map>
using namespace std;

int table[11] = {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89};


int main() {
    string input;
    cin >> input;
    
    const int n = input.size();
    set<string> res;
    unordered_map<char, int> map;
    
    string curr;
    for (int i=0; i<n; i++) {
        for (int j=1; j+i<= n; j++) {
            curr = input.substr(i, j);
            int count = 0;
            for (int k=0; k<curr.size(); k++) {
                if (map[curr[k]]==0) {
                    count++;
                    map[curr[k]] =1;
                }
            }
            for (int m=0; m<11; m++) {
                if (count == table[m]) {
                    res.insert(curr);
                    break;
                }
                else if (count <table[m])
                    break;
            }
            map.clear();
        }
    }
    
    set<string>::iterator it = res.begin();
    while (it!=res.end()) {
        cout << *it << endl;
        it++;
    }
    return 0;
}

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31/38

Welcome to Shanshan Blog!

DirectX Tutroial 1 First Window

Steps

hihoCoder 1082 然而沼跃鱼早就看穿了一切

Description

Input and Output

Codes

hihoCoder 1186 Coordinates

Description

Input and Output

Codes

hihoCoder 1148 2月29日

Description

Input and Output

Code

hihoCoder 1152 Lucky Substrings

Description

Input and Output

Code