將文本文件中的數據讀取到一個二維數組中，求出這些數據的中位數，然後求所有數據與中位數的差值的絕對值和，並求出和的最小值，最後把程序運行的結果寫到一個txt文件中

#include<iostream>
#include<fstream>
#include <string>
#include <sstream>
using namespace std;

ofstream outFile("final.txt");//寫到.txt文件中。

float zws(float **a, const int start1,const int end1,const int start2,const int end2)
{
	const int sum=(end1-start1+1)*(end2-start2+1);
	float* b = new float[sum + 1];
	float temp;
	int n = 0;
	float middle;
	float sum1 = 0.0,sum2 = 0.0;
	for(int i = start1;i<=end1;i++)//將二維數組裏的值放到一個一維數組中。
		for(int j = start2; j <= end2; j++, n++)
			b[n] = a[i][j];

	//下面對一維數組裏的值由小到大進行排序。
	for(int i = 0;i < sum;i++)
		for(int j = 0;j < sum-1 - i;j++)
			if(b[j] > b[j+1])
			{
				temp = b[j];
				b[j] = b[j+1];
				b[j+1] = temp;
			}

			//求中位數.
			if(sum%2 == 1)//奇數
			{
				middle = b[sum/2 + 1];
				outFile << "奇數中位數爲：" << middle << endl;
			}
			else		  //偶數
			{
				middle = (b[sum/2 + 1]+b[sum/2])/2;
				outFile << "偶數中位數爲：" << middle << endl;
			}

			for(int i = 0; i < sum; i++)
			{
				if(b[i] >= middle)
				{
					sum1 += 3*(b[i] - middle);
					sum2 += b[i] - middle;
				}
				else
				{
					sum1 += middle - b[i];
					sum2 += middle - b[i];
				}

			};

			//cout << "填挖土石所需的總費用爲：" << 900*sum1 << endl;
			outFile << "填挖土石所需的總費用爲：" << 900*sum1 << endl;

			float u = 900*sum2;
			delete b;
			return u;
}

int main()
{
	//動態申請二維數組
	float** a;
	a = new float*[31];
	for(int i=0; i<31; i++)
		a[i] = new float [51];

	ifstream fin("result.txt");//打開文件
	//讀入數字
	for(int i = 0; i <= 30; i++)
		for(int j = 0; j <= 50; j++)
			fin >> a[i][j];
	fin.close();

	float min = zws(a,0,26,0,19);
	float temp = 0.0;
	for(int i = 0; i <= 4; i++)
	{
		for(int j = 0; j <= 31; j++)
		{
			outFile << "****************************************" << endl;
			outFile << "i=" << i << "\t"<< "j=" << j << endl;
			outFile << "總土石方量爲 = " << zws(a, i, i+26, j, j+19) << endl;
			temp = zws(a, i, i+26, j, j+19);
			if(temp < min)
				min = temp;
		}
	}


	outFile << "***************************************" << endl;
	outFile << "總土石方量最小值爲：" << min << endl;

	system("pause");
	for(int i = 0; i <= 30; i++)
		delete[]a[i];
	delete []a;
	return 0;

}

輸出結果：