C++求出多元一次方程
阿新 • • 發佈:2019-02-05
注:
是n*n的等距矩陣,程式碼如下:
#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}, };//第四列是增廣矩陣 int i,j; n = 3; cout<<"輸入方程組介數:"; cout<<n<<endl; cout<<"輸入增廣矩陣:"<<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;//讓該行的所在列除以所在列的第一個元素,目的是讓首元素為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;//前後行數相減,使下一行或者上一行的首元素為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<<"方程組的解是:"<<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; } }