C++实现的求解多元一次方程示例

#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},{400,20,20},{900,30,};//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,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,4);
      }
      print(a,4);
    }
    print(a,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,4);
    }
  }
  print(a,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 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;
  }
}