[轉錄][轉錄][Ｃ++] 上機考 (7)

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※ [本文轉錄自 NTUBIME99-HW 看板]

作者: matrixjimmy (Jimmy) 看板: NTUBIME99-HW
標題: [轉錄][Ｃ++] 上機考 (7)
時間: Sun Dec 17 14:49:36 2006

※ [本文轉錄自 NTUBIME97-HW 看板]

作者: ychia (12/13~17 生機週!!) 看板: NTUBIME97-HW
標題: [Ｃ++] 上機考 (7)
時間: Wed Dec  8 00:25:28 2004

#include <iostream>

using std::cin;
using std::cout;
using std::endl;

#include <cmath>

int main()
{
    float a, b, c, D;  // ax^2 + bx + c, D = b^2 - 4ac

    cout << "這是一個計算二元一次方程式 ax^2 + bx + c = 0 的程式" << endl;
    cout << "請輸入 a : ";
    cin >> a;
    cout << "請輸入 b : ";
    cin >> b;
    cout << "請輸入 c : ";
    cin >> c;


    // 如果 a = 0, 則不須用到公式解, x = -c/b
    if ( a == 0 ) {
        if ( b != 0 )
            cout << "x = " << -1 * c / b << endl;
        else if ( b == 0 )  // 分母不能為零
            cout << "無解" << endl;
        else
            cout << "what condition?";
    }
    else {
        D = pow( b, 2) - 4 * a * c; // 判別式
        if ( D > 0 )
            cout << "x = " << ( -1 * b + sqrt(D) ) / ( 2 * a ) << " , "
                 << "x = " << ( -1 * b - sqrt(D) ) / ( 2 * a ) << endl;
        if ( D == 0 )
            cout << "x = " << ( -1 * b ) / ( 2 * a ) << " , "
                 << "x = " << ( -1 * b ) / ( 2 * a ) << endl;
        if ( D < 0 ) {  // 將負根號轉換成 i , 顯示分數
            D = -D;
            cout << "x = ( " << ( -1 * b ) << " + " << sqrt(D) << "i" << " )/" << ( 2 * a ) << " , "
                 << "x = ( " << ( -1 * b ) << " - " << sqrt(D) << "i" << " )/" << ( 2 * a ) << endl;
        }
    }

    return 0;
}




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