swift算法手记-5
发布时间:2020-12-14 07:20:27 所属栏目:百科 来源:网络整理
导读://// ViewController.swift// learn5//// Created by myhaspl on 16/1/23.// Copyright (c) 2016年 myhaspl. All rights reserved.//import Cocoaimport Foundationclass ViewController: NSViewController { override func viewDidLoad() { super.viewDidLo
// // ViewController.swift // learn5 // // Created by myhaspl on 16/1/23. // Copyright (c) 2016年 myhaspl. All rights reserved. // import Cocoa import Foundation class ViewController: NSViewController { override func viewDidLoad() { super.viewDidLoad() // Do any additional setup after loading the view. } override var representedObject: AnyObject? { didSet { // Update the view,if already loaded. } } private func comresult(inputnum:Double)->Double{ // 5*x^7-3*x^5+16*x^2+7*x+90=0 let myresult:Double = 5 * pow(inputnum,7) - 3 * pow(inputnum,5) + 16 * pow(inputnum,2) + 7 * inputnum + 90 return myresult } @IBOutlet weak var result: NSTextField! @IBAction func compute(sender: AnyObject){ // 5*x^7-3*x^5+16*x^2+7*x+90=0 // 二分法求一元方程的解,最大求解范围[-100000,100000] let trycount = 80 var accuracy: Double = 0.00000000000001 var answer: Double?=nil // 估计解范围 var leftbound:Double?=nil var rightbound:Double?=nil for var bound:Double=1;bound<10000000;bound*=10{ let leftres=comresult(-bound) let rightres=comresult(bound) if (leftres*rightres) < 0 { leftbound = (-bound) rightbound = bound break } } if (leftbound==nil || rightbound==nil){ return } //计算方程的解 for i in 1...trycount{ result.stringValue=String(i) let center=leftbound!+(rightbound!-leftbound!)/2 let leftres:Double=comresult(leftbound!) let rightres:Double=comresult(rightbound!) let centres:Double=comresult(center) if centres==0 { answer=center break } else if abs(rightbound!-leftbound!) < accuracy { answer=leftbound! break } else if leftres*centres<0{ rightbound=center } else if rightres*centres<0{ leftbound=center } } if let ans=answer{ //方程有解 result.stringValue="解:"+String(stringInterpolationSegment: ans)+" " result.stringValue += "解代入方程的值:"+String(stringInterpolationSegment:comresult(ans)) } } } 本博客所有内容是原创,如果转载请注明来源http://blog.csdn.net/myhaspl/(编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |