人教A版必须修读三高中数学演示课件1-1-2—1-1-3程序框图与算法的基本逻辑结构.ppt

开始 i=1 i100？ 是 输出S 结束 S=0 i=i+1 S=S+i 否 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 开始 i=1 结束 输出S 否 是 S=0 S=S+i i≤100？ i=i+1 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 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. 算法分析： Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. （3）控制条件：当“a300”时终止循环. （2）初始值：n=2005，a=200. 循环结构： Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 开始 n=2005 a=200 t=0.05a a=a+t n=n+1 a300？ 结束 输出n 是 否 程序框图： Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. （2）循环结构的程序框图各有两种形式. 循环结构的程序框图的基本特征： 小 结 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. * 湖南省长沙市一中卫星远程学校 * 湖南省长沙市一中卫星远程学校 §1.1.2 程序框图与算法的基本逻辑结构 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 课前复习 算法 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 知识探究（一）：算法的程序框图 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2. 我们将上述算法如下表示： 是 是 否 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. 算法的基本逻辑结构： Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 知识探究（二）：算法的顺序结构 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 第一步，输入三角形三条边的边长a，b，c. 第二步，计算 . 第三步，计算