题目1103：二次方程计算器

java实现：

import java.util.Scanner;
import java.io.IOException;
import java.io.FileReader;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
import java.util.Formatter;class Main
{public static final boolean DEBUG = false;public static void main(String[] args) throws IOException {Scanner cin;Pattern p = Pattern.compile("(\\+|-|=)");String s;double a, b, c;boolean bEqual, bMinus;int loc1;if (DEBUG) {cin = new Scanner(new FileReader("d:\\OJ\\uva_in.txt"));} else {cin = new Scanner(System.in); }while (cin.hasNext()) {s = cin.next();a = b = c = 0;bEqual = bMinus = false;loc1 = 0;String sub;Matcher m = p.matcher(s);while (m.find()) {sub = s.substring(loc1, m.start());if (sub.indexOf('x') == -1) {if ((!bEqual && !bMinus) || (bEqual && bMinus)) {if (sub.compareTo("") != 0) c += Integer.parseInt(sub);} else {if (sub.compareTo("") != 0)c -= Integer.parseInt(sub);}} else if (sub.indexOf('^') == -1) {if ((!bEqual && !bMinus) || (bEqual && bMinus)) {if (sub.length() == 1) b += 1;else b += Integer.parseInt(sub.substring(0, sub.length() - 1));} else {if (sub.length() == 1) b -= 1;else b -= Integer.parseInt(sub.substring(0, sub.length() - 1));}} else {if ((!bEqual && !bMinus) || (bEqual && bMinus)) {if (sub.length() == 3) a += 1;else a += Integer.parseInt(sub.substring(0, sub.length() - 3));} else {if (sub.length() == 3) a -= 1;else a -= Integer.parseInt(sub.substring(0, sub.length() - 3));}}if (m.group().compareTo("=") == 0) bEqual = true;else if (m.group().compareTo("-") == 0) bMinus = true;else bMinus = false;loc1 = m.end();}sub = s.substring(loc1, s.length());if (sub.indexOf('x') == -1) {if ((!bEqual && !bMinus) || (bEqual && bMinus)) {if (sub.compareTo("") != 0) c += Integer.parseInt(sub);} else {if (sub.compareTo("") != 0) c -= Integer.parseInt(sub);}} else if (sub.indexOf('^') == -1) {if ((!bEqual && !bMinus) || (bEqual && bMinus)) {if (sub.length() == 1) b += 1;else b += Integer.parseInt(sub.substring(0, sub.length() - 1));} else {if (sub.length() == 1) b -= 1;else b -= Integer.parseInt(sub.substring(0, sub.length() - 1));}} else {if ((!bEqual && !bMinus) || (bEqual && bMinus)) {if (sub.length() == 3) a += 1;else a += Integer.parseInt(sub.substring(0, sub.length() - 3));} else {if (sub.length() == 3) a -= 1;else a -= Integer.parseInt(sub.substring(0, sub.length() - 3));}}if (b * b - 4 * a * c < 0) {System.out.println("No Solution");} else {double x1 = (-b + Math.sqrt(b * b - 4 * a * c)) / (2 * a);double x2 = (-b - Math.sqrt(b * b - 4 * a * c)) / (2 * a);if (x1 > x2) {double tmp = x1; x1 = x2;x2 = tmp;}Formatter fmt= new Formatter();fmt.format("%.2f %.2f", x1, x2);System.out.println(fmt);}}}
}

C++实现：

#include <iostream>
#include <fstream>
#include <string>
#include <cmath>
#include <algorithm>
#include <iomanip>using namespace std;int main()
{string s;string sop = "+-=";size_t loc1, loc2;float a, b, c;bool bEqual, bMinus;#ifndef ONLINE_JUDGEifstream cin("d:\\OJ\\uva_in.txt");
#endif // ONLINE_JUDGEwhile (cin >> s) {loc1 = -1;loc2 = 0;bEqual = false;bMinus = false;a = b = c = 0;while (loc2 < s.length()) {loc2 = s.find_first_of(sop, loc1 + 1);if (loc2 == string::npos) {loc2 = s.length();}string sub = s.substr(loc1 + 1, loc2 - loc1 - 1);//cout << "sub:" << sub << endl;if (sub.find('x') == string::npos) {if ((!bEqual && !bMinus) || (bEqual && bMinus))c += atoi(sub.c_str());elsec -= atoi(sub.c_str());} else if (sub.find('^') == string::npos) {if ((!bEqual && !bMinus) || (bEqual && bMinus)) {if (sub.length() == 1) {b += 1;} else {b += atoi(sub.c_str());}} else {if (sub.length() == 1) {b -= 1;} else {b -= atoi(sub.c_str());}}} else {if ((!bEqual && !bMinus) || (bEqual && bMinus)) {if (sub.length() == 3) {a += 1;} else {a += atoi(sub.c_str());}} else {if (sub.length() == 3) {a -= 1;} else {a -= atoi(sub.c_str());}}}loc1 = loc2;if (loc2 < s.length() && s.at(loc2) == '-') bMinus = true;else if (loc2 < s.length() && s.at(loc2) == '=') bEqual = true;else bMinus = false;//cout << "a:" << a << " b:" << b << " c:" << c << endl;}if (b * b - 4 * a * c < 0) {cout << "No Solution" << endl;} else {float result1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);float result2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);if (result1 > result2) swap(result1, result2);cout.precision(2);cout.setf(ios::fixed);cout << result1 << " " << result2 << endl;}}return 0;
}

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