[程勇电工技术教材公共课件：模块四.ppt

Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 算法与程序框图 算 法 程序框图 算法的三种基本逻辑 结构和框图表示 循环结构 条件分支结构 顺序结构 知识回顾： Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.  四种图框类型 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 语句A 语句B 知识再现 顺序结构 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 条件结构 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 开始 程序框图 x0? 是 y=(x+2)2 否 x=0? 是 y=4 否 输出y 结束 输入x y=(x-2)2 讲义第五页例2 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 开始 A=1 n=1 A=A-1 n=n+1 n5? 是 结束 否 输出 A 如图所示的程序框图，运行该 程序，输出的第3个数是——— -1 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 求1+2+3+…..+100的算法 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 思考:将步骤A和步骤B交换位置，结果会怎样？能达到预期结果吗？为什么？要达到预期结果，还需要做怎样的修改？ 答：达不到预期结果；当i = 100时，退出循环，i的值未能加入到S中；修改的方法是将判断条件改为i100，i的初值也应相应改变 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 练习巩固 1、设计一算法，求积:1×2×3×…×100，画出流程图 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2、 对任意正整数n, 的值,并画出程序框图. 开始 输入正整数n 输出S 结束 S=0 i=1 S=S+1/i i=i+1 in? 否 是 设计一个算法求 练习巩固 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2、设计一算法输出1~1000以内能被3整除的整数 算法： S1：确定i的初始值为0； S2：判断i是否等于1000，若是则程序结束，否则进入S3； S3：使i增加1，判断i是否能被3整除，若能输出i，