现场编码题目-2

现场编码题目2：
某集团有三个单位A、B、C，三个单位总人数有900人,A单位人数的比B单位人数多320人,C单位人数平方比A单位人数少200人,请问A、B、C三个单位人数各自多少?
编码算法规定:通过三层循环穷举方法，同时,注意循环算法的优化。正确答案:A=600,B=280,C=20,请用编码求解。

 声明:下方的参考代码并不是最优解法。

参考代码Python版:

bFinished=False
for A in range(1,901):
    for B in range(1,901):
        for C in range(1,901):
            if(A+B+C==900 and A-B==320 and A-C*C==200):
                print(A,B,C)
                bFinished=True
                break
        if(bFinished==True):
            break
    if(bFinished==True):
        break
print("over")

 参考代码C#版：

using System;

namespace Case2
{
    class Program
    {
        static void Main(string[] args)
        {
            bool bFinished = false;
            for (int A = 1; A <= 900; A++)
            {
                for (int B = 1; B <= 900; B++)
                {
                    for (int C = 1; C <= 900; C++)
                    {
                        if (A + B + C == 900 && A - B == 320 && A - C * C == 200)
                        {
                            Console.WriteLine("A:{0},B:{1},C:{2}", A, B, C);
                            bFinished = true;
                            break;
                        }
                    }
                    if (bFinished == true)
                        break;                
                }
                if (bFinished == true)
                    break;
           }
            Console.WriteLine("over");
        }
    }
}

 参考代码 C++:

#include <iostream>
using namespace std;

int main()
{
    bool bFinished = false;
    for (int A = 1; A <= 900; A++)
    {
        for (int B = 1; B <= 900; B++)
        {
            for (int C = 1; C <= 900; C++)
            {
                if (A + B + C == 900 && A - B == 320 && A - C * C == 200)
                {
                    cout << "A:"<<A << "  B:"<<B << "  C:"<<C<<endl;
                    bFinished = true;
                    break;
                }
            }
            if (bFinished == true)
            {
                break;
            }
        }
        if (bFinished == true)
        {
            break;
        }
    }  
}