Search - max - CodeBus

Search - max - List

[Other Games] Max7_SDK DL : 0: max插件用的... 做游戏可以参考的啊啊啊啊 -max plug-ins used to do the game ... can refer to the ah ah ah ah
Date : 2025-12-28 Size : 9.99mb User :
[Other Games] tutorials DL : 0: 3ds max for ogre的输出教程．开发网络游戏有帮助-3ds max for ogre tutorials output. Help the development of online games
Date : 2025-12-28 Size : 3.73mb User : lyj
[Other Games] max DL : 0: max低版本打来高版本文件演示用了以后猜知道-high-low version of max call demo version of a document
Date : 2025-12-28 Size : 389kb User : li
[Other Games] healthbar DL : 0: Just a simple health bar for an RPG or so. First text box is min health, second is max.
Date : 2025-12-28 Size : 1kb User : scorpionz
[Other Games] zhuxian-max-scene DL : 0: 诛仙网游河阳城的场景文件。.max .dds完整。-the scene file including whole required .max and dds files.
Date : 2025-12-28 Size : 7.6mb User : 王内秀
[Other Games] juzhenxiangcheng DL : 0: 编写程序，可以实现m*n矩阵和n*p矩阵相乘。m,n,p均小于10，矩阵元素为整数。分析：首先我们可以根据题意写出函数头。可以定为void MatrixMutiply(int m,int n,int p,long lMatrix1[MAX][MAX],long lMatrix2[MAX][MAX],long lMatrixResult[MAX][MAX])，其中lMatrix1和lMatrix2分别是输入的m*n矩阵和n*p矩阵，lMatrixResult是输出的m*p矩阵。因为m，n和p都是未知量，要进行处理的矩阵大小是变量。但我们可以定义比较大的二维数组，只使用其中的部分数组元素。矩阵相乘的算法比较简单，输入一个m*n矩阵和一个n*p矩阵，结果必然是m*p矩阵，有m*p个元素，每个元素都需要计算，可以使用m*p嵌套循环进行计算。根据矩阵乘法公式：可以用循环直接套用上面的公式计算每个元素。嵌套循环内部进行累加前，一定要注意对累加变量进行清零。-Write a program, you can achieve the m* n matrix and n* p matrix multiplied. m, n, p are less than 10, the matrix elements are integers. Analysis: First of all, we can write the function header according to the meaning of the questions. As void MatrixMutiply (int m, int n, int p, long lMatrix1 [MAX] [MAX], long lMatrix2 [MAX] [MAX], long lMatrixResult [MAX] [MAX]) of which lMatrix1 and lMatrix2, are input m* N matrix and n* p matrix, lMatrixResult is the output of m* p matrix. M, n and p are unknown quantity, the size of the matrix to be processed is variable. However, we can define a large two-dimensional arrays, which use only part of the array elements. M* p nested relatively simple matrix multiplication algorithm, the input of an m* n matrix, and an n* p matrix, the inevitable result is a matrix of m* p, there are m* p elements, each element needs to be calculated and can be used loop calculation. Matrix multiplication formula: cycle directly apply the above formula to calculate ea
Date : 2025-12-28 Size : 11kb User : 郑雯雯
[Other Games] 3 DL : 0: max文件，人体模型，好用实在，西部酷男-max file
Date : 2025-12-28 Size : 19kb User : chou
[Other Games] fuzi DL : 0: 3ds max 做的游戏装备，斧子，用放样的效果做的，不懂的别下-3ds max to do game equipment, ax
Date : 2025-12-28 Size : 20kb User : 刘志军
[Other Games] 别吃我 DL : 0: 一个圆将几个文字吃掉，里面具有设计报告源文件和
Date : 2013-12-01 Size : 1.97mb User : 786571770
[Other Games] maxCamera DL : 0: unity3d中仿造max的拖拽，拉距，旋转方式代码-unity3d in counterfeit max drag, pull away, rotation mode code
Date : 2025-12-28 Size : 1kb User : chenhao
[Other Games] 14 DL : 0: 求解二维数组的最大/最小元素 -min or max
Date : 2025-12-28 Size : 1kb User : pdsun
[Other Games] newbur DL : 0: Utiliza arreglos con una constante (MAX) que indica el número máximo de números a ordenar.
Date : 2025-12-28 Size : 1kb User : tigrerodrigo
[Other Games] Pacman2-Izabella,Max DL : 0: Implementation of multiagent of the pacman project
Date : 2025-12-28 Size : 318kb User : bebellamc