高一数学必修三序程框图与算法的基本逻辑结构.ppt

Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 算法步骤有明确的顺序性，而且有些步骤只 有在一定条件下才会被执行，有些步骤在一定条 件下会被重复执行.算法可以用自然语言来描述， 但为了使算法的程序或步骤表达得更为直观、准 确，我们更经常地用图形方式来表示它。 程序框图又称流程图，是一种用程序框、流 程线及文字说明来表示算法的图形 一个程序框图包括以下几部分：表示相应操 作的程序框；带箭头的流程线；程序框外必要的 文字说明。 1.程序框图 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 图形符号名 称功 能 终端框（起止框） 一个算法的起始和结束 输入、输出框 一个算法输入和输出的信息 处理框（执行框） 赋值、计算 判断框 判断某一条件是否成立，出 口成立标“是”不成立标“ 否” 流程线 连接程序框 连接点 连接程序框图的两部分 或 构成程序框的图形符号及其作用 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. N不是质数 是 否 判断整数n（n2)是否为质数 程序框图范例： 开始 输入n i=2 求n除以i的余数r i的值增加1，仍用i表示 in-1或r=0? r=0? N是质数 结束 是否 设n是一个大于2的整数 一般用i=i+1表示 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 开始 求n除以i的余数r i=2 输入n in-1？ r=0？ 输出“n 是质数” 是 i的值增加1, 仍用i表示 否 输出“n不 是质数” 是 结束 否 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 学习这部分知识的时候，要掌握各个图形的 形状、作用及使用规则，画程序框图的规则如下： 1、使用标准的图形符号。 2、框图一般按从上到下、从左到右的方向画。 3、除判断框外，大多数流程图符号只有一个进入 点和一个退出点。判断框是具有超过一个退出 点的唯一符号。 4、判断框分两大类，一类判断框是“是”与“否”两分 支的判断，而且有且仅有两个结果；另一类是 多分支判断，有几种不同的结果。 5、在图形符号内描述的语言要非常简练清楚。 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 练 习 1、下列关于程序框图的说法正确的是 （ ） A、程序框图是描述算法的语言 B、程序框图可以没有输出框，但必须要有输入框给变量赋值 C、程序框图可以描述算法，但不如自然语言描述算法直观 D、程序框图和流程图不是一个概念 2、下列功能“ ”没有功能的是 （ ） A、赋值 B、计算 C、判断 D、 以上都不对 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 开始 输入n i=2 求n除以i的余数r i的值增加1，仍用i表示 in-1或r=0? r=0? N不是质数 N是质数 结束 是 否 是否 开始 结束 求n除以i的余数r i的值增加1，仍用i表示 in-1或r=0? 是 否 r=0? N不是质数 N是质数 是否 输入n i=2 顺序结构 循环结构 条件结构 2.算法的三种基本逻辑结构：顺序结构、条件结构、循环结构。 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 求n除以i的余数r i的值增加1，仍用i表示 in-1或r=0? 是 否 r=0? N不是质数 N是质数 是否 输入n i=2 顺序结构 循环结构 条件结构 算法千差万别，但都是由这 三种基本逻辑结构构成的. 你能说出这三种基本逻辑结构的特点吗？ 条件结构与循环结构有什么区别和联系？ Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. （1）顺序结构 顺序结构是最简单的算法结构，语句与语句 之间，框与框之间是按从上到下的顺序进行的， 它是由若干个依次执行的处理步骤组成的，它是 任何一个算法都离不开的一种基本算法结构。 顺序结构在程序框图中的体现就是用流程线 将程序框自上而下地连接起来，按顺序执行算法 步骤。 步骤n 步骤n1 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 例3、已知一个三角形的三边分别为a、b、c，利用海伦公式设 计一个算法，求出它的面积，并画出算法的程序框图。 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 开始 输入a,b,c 结束 p S 输出S Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 例4、已知两个变量A和B的值，试设计一个交 换这两个变量的值的算法，并画出程序框图 。 第一步、输入A、B 第二步、令X=A 第三步、令A=B 第四步、令B=X 第五步、输出A、B Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 练 习 1、已知摄氏温度C与华氏温度F之间的关系为 F=1.8C+32。设计一个由摄氏温度求华氏温度 的算法，并画出相应的程序框图。 算法步骤： 第一步：输入摄氏温度C； 第二步：计算1.8C+32，并 将这个值记为华氏温度F； 第三步：输出华氏温度F。 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 练 习 2、已知变量A、B、C的值，试设计一个算法 程序框图，使得A为B的值，B为C的值，C为A 的值。 第一步、输入A、B、C 第二步、令X=A 第三步、令A=B 第四步、令B=C 第五步、令C=X 第六步、输出A、B、C的值 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. （2）条件结构 在一个算法中，经常会遇到一些条件的判断， 算法的流程根据条件是否成立有不同的流向.条件 结构就是处理这种过程的结构. 分类是算法中经常发生的事情，条件结构的 主要作用就是表示分类. 条件结构可用程序框图表示为下面两种形式. 步骤A步骤B 满足条件？ 否 是 步骤A 满足条件？ 否 是 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 例4 任意给定3个正 实数,设计一个算法,判 断分别以这3个数为三 边边长的三角形是否存 在.画出这个算法的程 序框图. 条件结构 算法步骤如下： 第一步，输入3个正实数 a,b,c. 第二步，判断a+bc，a+cb ，b+ca是否同时成立. 若是，则存在这样的三角形; 否则，不存这样的三角形. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 条件结构 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 例5 设计一个求解一元二次方程 的算法，并画出程序框图表示. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 开 始 输入a，b，c = 4ac 0? 0? 输出 , 结 束 方程无实数根输出x 否 是 是 否 例5程序框图也可设计为 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 例6、设计一个算法计算分段函数 ， 的函数值，并画出程 序框图。 第一步、输入x 第二步、判断“x3,则 费用为m= 5（x-3）1.2=1.2x+1.4; 若x3,则费用为m5. 第三步，输出m. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 开始 输入a1，b1，c1， a2 。b2， c2 输出“x= ”; ，“y= ”; a1b2a2b10？ x=(c2b1 c1b2)/(a1b2 a2b1) y=(c2b1 c1b2)/(a1b2 a2b1) 结束 N Y 输出“输入数据不合题意 P.20 习题1.1B组第1题 算法步骤： 第一步，输入a1，b1，c1， a2，b2，c2. 第二步，计算 x=(c2b1 c1b2)/(a1b2 a2b1) y=(c2b1 c1b2)/(a1b2 a2b1) 第三步，输出x，y Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. （3）循环结构 循环结构指的是按照一定的条件反复执行的某些算法步骤. 反复执行的步骤称为循环体. 循环体 满足条件？ 否 是 循环体 满足条件？ 否 是 执行一次循环体后，对条件进行 判断，如果条件不满足，就继续执行 循环体，直到条件满足时终止循环. 在每次执行循环体前，对条件进行 判断，当条件满足，执行循环体，否则 终止循环. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 例7 设计一个计算1+2+3+100的值的算法，并画 出程序框图. 第一步：确定首数a，尾 数b，项数n； 第二步：利用公式“S=n (a+b) /2”求和； 第三步：输出求和结果。 算法1： 开始 结束 输入a,b,n S=n (a+b) /2 输出S Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 开始 i=1 S=0 i=i+1 S=S+i i100? 输出S 结束 否 是 例7 设计一个计算 1+2+3+100的值的算法， 并画出程序框图. 第1步，011. 第2步，123. 第3步，336. 第4步，6410. 第100步，49501005050. 算法2： 第一步，令i1，S0. 第二步，若i 100成立，则执 行第三步;否则，输出S，结束算法. 第三步，SSi. 第四步，i=i+1,返回第二步.当型循环结构 （1）确定循环体：i=i+1 s=s+i (2)初始化变量：i=1 s=0 （3）循环控制条件：i100 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 结束 s = s+i i=i+1 i100? 输出S 否 是 i=0；S=0 开始 结束 输出S i=0；S=0 开始 S=s+i i=i+1 i100? 否 是 循环结构 直到型结构当型结构 例7 设计一个计算1+2+3+100的值的算法，并画 出程序框图. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. s = s + i i=i+ 1 解决方法就是加上一个判断， 判断是否已经加到了100，如果加到 了则退出，否则继续加。 直到型结构 当型结构 s=s+i i=i+1 是 否 s =s+i i=i+1 否 是 i100? i100? 请填上判断的条件。 在解题的过程中，用累加变量S表示 每一步的计算结果，即把S+i的结果仍记 为S，从而把第i步表示为S=Si,其中S 的初始值为0，i依次取1，2，100. 由于i同时记录了循环的次数，所以也称 为计数变量. 循环结构中都有一个计数变量和累加变量， 计数变量用以记录循环次数，同时它的取值还 用于判断循环是否终止，累加变量用于输出结 果，累加变量和计数变量一般是同步执行的， 累加一次，计数一次. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 例8 某工厂2005年的年生产总值为 200万元，技术革新后预计以后每年的年 生产总值都比上一年增长5.设计一个程 序框图，输出预计年生产总值超过300万 元的最早年份. 算法步骤： 第一步，输入2005年的年生产总值. 第二步，计算下一年的年生产总值. 第三步，判断所得的结果是否大于 300.若是，则输出该年的年份;否则，返 回第二步. 结束 开始 输出n a=200 t=0.05a a=a+t n=n+1 a300? Y n=2005 N （1）确定循环体：设a为某年的年生产 总值，t为年生产总值的年增长量，n为 年份，则循环体为t=0.05a,a=a+t,n=n+1. (2)初始化变量： n=2005， a=200. （3）循环控制条件： a300 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 结束 开始 输入n a=200 t=0.05a a=a+t n=n+1 a300? Y n=2005 N 结束 开始 输入n a=200 t0.05a aa+t nn+1 a300? N n=2005 Y 直到型当型 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 小结 1、循环结构的特点 2、循环结构的框图表示 3、循环结构该注意的问题 避免死循环的出现，设置好进入（结束 ）循环体的条件。 当型和直到型 重复同一个处理过程 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty