本文实例讲述了C++实现的求解多元一次方程。分享给大家供大家参考,具体如下:
注:这里计算的是n*n的等距矩阵,代码如下:
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#include<iostream> #include<math.h> #include<fstream> #include<stdlib.h> using namespace std; void print( double (*pArray)[4], int iWidth, int iHigh); void main(){ int n,m; double a[3][4] = { {100, 10, 1, 10}, {400, 20, 1, 20}, {900, 30, 1, 10}, }; //第四列是增广矩阵 int i,j; n = 3; cout<< "输入方程组介数:" ; cout<<n<<endl; cout<< "输入增广矩阵:" <<endl; for (i = 0; i < n; i++){ for (j = 0; j < n + 1;j++){ cout<<a[i][j]<< " " ; } cout<<endl; } for (j = 0; j < n; j++){ double max = 0; double imax = 0; for (i = j; i < n; i++){ if (imax < fabs (a[i][j])){ imax = fabs (a[i][j]); max = a[i][j]; //得到各行中所在列最大元素 m = i; } } if ( fabs (a[j][j]) != max) { double b = 0; for ( int k = j;k < n + 1; k++){ b = a[j][k]; a[j][k] = a[m][k]; a[m][k] = b; } } print(a, 3, 4); for ( int r = j;r < n + 1;r++){ a[j][r] = a[j][r] / max; //让该行的所在列除以所在列的第一个元素,目的是让首元素为1 } print(a, 3, 4); for (i = j + 1;i < n; i++){ double c = a[i][j]; if (c == 0) continue ; for ( int s = j;s < n + 1;s++){ double tempdata = a[i][s]; a[i][s] = a[i][s] - a[j][s] * c; //前后行数相减,使下一行或者上一行的首元素为0 print(a, 3, 4); } print(a, 3, 4); } print(a, 3, 4); } for (i = n - 2; i >= 0; i--){ for (j = i + 1;j < n; j++){ double tempData = a[i][j]; double data1 = a[i][n]; double data2 = a[j][n]; a[i][n] = a[i][n] - a[j][n] * a[i][j]; print(a, 3, 4); } } print(a, 3, 4); cout<< "方程组的解是:" <<endl; for ( int k = 0; k < n; k++){ cout<< "x" <<k<< " = " <<a[k][n]<<endl; } } void print( double (*pArray)[4], int iWidth, int iHigh) { std::cout<< "Array: " << "\n" ; for ( int i = 0; i < iWidth; i++){ for ( int j = 0; j < iHigh;j++){ cout<<pArray[i][j]<< " " ; } cout<<endl; } } |
PS:这里再为大家推荐几款计算工具供大家进一步参考借鉴:
科学计算器在线使用_高级计算器在线计算:https://tool.zzvips.com/t/jsq/
希望本文所述对大家C++程序设计有所帮助。
原文链接:http://blog.csdn.net/ganpengjin1/article/details/22959923