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Description: 三角形逼近球体演示
原理：先用正三角形逼近绘出20面体，然后递归分割，把每个面分割为4个三角形(递归次数可定),此以来逼近球体。-spherical approximation triangle demonstration tenets : first approximation using regular triangle charted 20 Verses, and then recursive segmentation, each side of the divided four triangles (recursive number may be), since this sphere approximation.
Platform: |
Size: 8408 |
Author: 李为民 |
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Description: 一个杨辉三角形和判断显示两个数之间的素数-a triangle and judgment shows the number of two prime numbers
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Size: 1999 |
Author: 李星明 |
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Description: 利用二十面体逼近球体的方法，不断的产生新的三角形，其中depth表示二十面体中的每个三角形要进行多少次的四等分。-use icosahe spherical approximation method, and constantly create new to the triangle. which said the 20-depth body of each triangle to the number of occasions in the four sub.
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Size: 22933 |
Author: mark |
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Description: 给出了二次函数的Julia集分形图的概念及逃逸时间算法绘制复杂分形图的基本原理,对Julia集给出了严格的数学定义.逃逸时间算法即取定迭代次数界限N,经N次迭代后,若x点仍在给定的区域内,则认为x是分形A中的点 否则x不是分形A中的点.该算法同样适用于Mandelbrot集、Sierpinski三角形等其他复杂分形图.试验表明,该算法绘制的Julia集分形图准确有效、优美清晰,算法简单实用.
-given quadratic function of the Julia set fractal images and the concept of escape time algorithm mapping complex fractal graph - The principle of Julia Sets is a strict mathematical definition. Escape from time algorithm that will limit the number of iterations N, by the N-th iteration, if x is still point to the region, think x A fractal is the point otherwise x A fractal is not the point. It is also applicable to Mandelbro t set, Sierpinski triangle and other complex fractal images. Test shows that The algorithm mapping the Julia set fractal images accurately and effectively, a beautiful clear, simple and practical algorithm.
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Size: 1136 |
Author: 为听 |
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Description: 用户指定顶点，系统判定并分解该多边形为不重叠的多个三角形，并使所有三角形的边长和最小-vertex specified users, the system and determine the polygon decomposition of the number of non-overlapping triangles, make all the triangle and the length of the smallest
Platform: |
Size: 448843 |
Author: Li Xia |
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Description: some simple but time wasting math logic
一些不难但麻烦（适合用电脑作）的数学函数的源码
1.Test for Prime numbers
2.Find the factorial of a number
3.Find the triangle(total sum) of a number
4.Reverse every integer of a number
5.Find any power of any value
6.Find the greatest common factor of 2 -some simple but time wasting some math logic difficult but trouble (suitable for use as computer) mathematical function source 1.Test for Prime number 's 2.Find the factorial of a number 3.Find the tri angle (total sum) of a number in every 4.Reverse teger of a number 5.Find any power of any value 6. Find the greatest common factor of two
Platform: |
Size: 1580 |
Author: jeffery |
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Description: some simple but time wasting math logic
一些不难但麻烦（适合用电脑作）的数学函数的源码
1.Test for Prime numbers
2.Find the factorial of a number
3.Find the triangle(total sum) of a number
4.Reverse every integer of a number
5.Find any power of any value
6.Find the greatest common factor of 2 -some simple but time wasting some math logic difficult but trouble (suitable for use as computer) mathematical function source 1.Test for Prime number 's 2.Find the factorial of a number 3.Find the tri angle (total sum) of a number in every 4.Reverse teger of a number 5.Find any power of any value 6. Find the greatest common factor of two
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Size: 1024 |
Author: jeffery |
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Description: 这是一个符号三角形问题的求解。第一行有n个符号。符号可以为“＋”或“－”，以下每一行的符号由上行得到，2个同号下面都是“+”，2个异号下面都是“－”，计算有多少个不同的符号三角形，使其所含的“＋”和“－”的个数相同。用户可以说输入多组测试数据n，以0为结束标志。-This is a symbolic triangle of solving the problem. There are n the first line of symbols. Symbols for or - , the following symbols from each line up to be, with the No. 2 the following are+ , 2 months following its all different - to calculate how many different symbols triangle so that it contained and - the same number. Users can be said that the test data-input multiple-group n, to 0 to end the signs.
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Size: 11264 |
Author: 谭晓彤 |
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Description: 求解符号三角形问题.给定一行符号(比如+和-),按照同号为+异号为-的原则计算第二行直至下面的行,最后的一行只有一个符号,要求统计各个符号的个数.-Problem solving triangle symbols. Given his symbols (such as+ And-), in accordance with the same number is+ Different number- the principle of calculation of the second line until the following line, the last line there is only one symbol, all symbols requested statistics number.
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Size: 10240 |
Author: Lotus Year |
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Description: 水仙花数是三位数，它的各位数字的立方和等于这个三位数本身，例如371—33+73+1 3，则371就是一个水仙花数。编程找出所有水仙花数。-Narcissus is the three-digit number, it figures that the three-digit cube and equal to itself, for example, 371-33+73+1 3, the 371 is a few daffodils. Programming to identify the number of all daffodils.
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Size: 1024 |
Author: yanglingfei |
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Description: 主要功能 ： 1、画图，并有坐标尺
2、能够随意的放大缩小（编写两个函数，分别对应X,Y粥的缩放倍数）
3、一个记录集(或链表)，记录当前画了多少图形，分别是什么图形，颜色，状态等
4、一个枚举类型，枚举所有本控件能画的图形（长方形，正方形，三角形 及自定义(即鼠标随意画)等）
5、能通过鼠标删除所画图形，并能移动图形，填充图形等
6、能够导出位图文件(bmp)
7、能够导出其他文件并能够导入显示-Main functions: 1, drawing and sitting ruler 2, is able to zoom in (the preparation of the two functions, corresponding to X, Y of the scaling multiplier porridge) 3, a record set (or list), drew a record number of current graphics, respectively, what is the graphics, color, state 4, an enumeration type, an enumeration of all the controls to draw graphics (rectangular, square, triangle and self-definition (that is, free to draw the mouse), etc.) 5, the mouse can delete painted graphics, and mobile graphics, graphics-filled 6, can export bitmap file (bmp) 7, can be exported into other documents and be able to show
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Size: 192512 |
Author: 老夫子 |
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Description: 运用glRotatef围绕远点画多个三角形-GlRotatef painting using a number of triangles
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Size: 998400 |
Author: dd |
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Description: 给定一个由n行数字组成的数字三角形如下图所示。
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
试设计一个算法，计算出从三角形
的顶至底的一条路径，使该路径经过的数字总和最大。-Given an n-line numbers from the number of triangles as shown below. 7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
trial design an algorithm to calculate the triangle from the top to the bottom of a path, so that the sum of the path through the largest number.
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Size: 1024 |
Author: bingyu |
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Description: 绘制若干OpenGL基本体素（三角形、三角形带、四边形、四边形带）构成的球体，注意球体绘制与球体划分的区别。
可以控制改变球的数量和球的体积，来改变基本体素的数量以及顶点的数量。
可以通过改变glPolygonMode，切换是否填充所绘制的基本体素。
利用函数glutGet(GLUT_ELAPSED_TIME)（returns the time in milliseconds since glutInit or the first invocation of the function）统计绘制时间
-Basic OpenGL rendering voxel number (triangle, triangle strips, quadrilateral, quadrilateral zone) constitutes the sphere, drawing attention to the sphere and the sphere by the difference. The number can control the ball and the ball change the volume, to change the basic number of voxels and the number of vertices. You can change the glPolygonMode, is populated by switching the basic voxel rendering. Using function glutGet (GLUT_ELAPSED_TIME) (returns the time in milliseconds since glutInit or the first invocation of the function) Statistics drawing time
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Size: 3072 |
Author: hongyi |
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Description: 输入一个数字，输入相应菱形的行数，注意不要输入太大的数值，以免显示屏范围超出-Enter a number, enter the appropriate diamond line number, be careful not to enter too much value, so that the display beyond
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Size: 1024 |
Author: 容永志 |
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Description: Print a number-triangle
Write a program that asks a user to input an integer number that is greater than 0 and
smaller than 10, and then outputs a number-triangle. For example, if the input is 2, the
number-diamond looks like this:
1
121
If the input is 5, the number-diamond looks like this:
1
1
121
12321
1234321
123454321
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Size: 1024 |
Author: xyzhu |
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Description: 第一，DDS模块是一个比较常用的用数字方式实现模拟信号的方法，以前一直只用了频率控制，这一次还通过深入理解用上了相位控制，从这个角度来讲，可以用FPGA小菜一碟的实现频率和相位可控的多通道SPWM波，然后再去外加上RC滤波电路和运放电路就可以实现可控正弦波。
第二，这里的DDS模块还有产生一个可逆计数器的计数使能时钟和方向控制时钟，需要具体说说的是，如果你输出的正弦值是8位的，那么你的计数器的计数范围是在0---255---0，如果你输出的正弦值是9位的，那么你的计数器的计数范围是在0---511---0。还有，每此计数变化后（一个“数字”三角波产生），DDS中只输出一个正弦值，进行比较。-First, the DDS module is a more commonly used in digital way to realize analog signal, the method of used in frequency control, this time also by deep understanding of using phase control, from this perspective, you can use the FPGA implementation of a piece of cake of frequency and phase controlled multichannel SPWM wave, and then to plus the RC filter circuit and the op-amp circuit can realize controllable sine wave.
Second, the DDS module and create a reversible counter count can make the clock and direction control clock, need specific about is that if you output sine value is 8 bits, so is the scope of your counter counts in 0-255-0, if your output sine value is nine, so is the scope of your counter counts in 0-511-0. Also, every change after the count (a "number" triangle wave), only in the DDS output a sine value, the comparison
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Size: 162816 |
Author: 张雪亮 |
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Description: The Triangle
时间限制：1000 ms | 内存限制：65535 KB
难度：4
描述
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
输入
Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.
输出
Your program is to write to standard output. The highest sum is written as an integer.
样例输入
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
样例输出
30-The Triangle
Time Limit: 1000 ms | Memory Limit: 65535 KB
Difficulty: 4
Description
7
38
810
2744
45265
(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
Enter
Your program is to read from standard input The first line contains one integer N:... The number of rows in the triangle The following N lines describe the data of the triangle The number of rows in the triangle is> 1 but <= 100 . The numbers in the triangle, all integers, are between 0 and 99.
Export
Your program is to write to standard output. The highest sum is written as an integer.
Sample Input
5
7
38
810
2744
45265
Sample Output
30
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Size: 1024 |
Author: Qinzhen |
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Description: 输入三个数，判断这三个数能不能构成直角三角形，思路是先求出最大数，再判断是否符合勾股定理-Enter the three count to determine the number of these three can not constitute a right triangle, the idea is to find the maximum number, and then determine whether the Pythagorean theorem
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Size: 559104 |
Author: lucky |
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Description: Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input
The first line of the input contains integer number T, the number of test cases in this test. Then following T lines of input containing four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output
For each test case, output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
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Size: 1067 |
Author: fujianyd@163.com |
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