Use the kramo law to solve the n-element linear equations.

Use the kramo law to solve the n-element linear equations.

Next, use the kramo rule to solve the n-element linear equations.

Code:

1 # include <iostream> 2 # include <cmath> 3 using namespace std; 4 void appDescribe (); // program description 5 double valDet (double * detPtr, int rank ); // solve the determinant 6 class LinearEquations 7 {8 public: 9 LinearEquations (); // construct the equations 10 void sol (); // solve 11 void showAns (); // display solution 12 private: 13 int rank; // Order 14 double * parPtr; // coefficient determinant 15 double * conPtr; // constant item 16 double * ansPtr; // solution 17}; 18 LinearEquations: LinearEquatio Ns () 19 {20 cout <"Number of input equations:"; 21 cin> rank; 22 int rank2 = rank * rank; 23 parPtr = new double [rank2]; 24 conPtr = new double [rank]; 25 cout <endl <"determinant of input coefficient:" <endl; 26 for (int I = 0; I <rank2; I ++) 27 cin> parPtr [I]; 28 cout <endl <"input constant:" <endl; 29 for (int I = 0; I <rank; I ++) 30 cin> conPtr [I]; 31} 32 void LinearEquations: sol () 33 {34 int rank2 = rank * rank; 35 double det = valDet (parPtr, rank); 36 ansPtr = new doub Le [rank]; 37 double * tempParptr = new double [rank2]; 38 double tempDet; 39 if (det = 0) 40 {41 ansPtr = NULL; 42 return; 43} 44 for (int I = 0; I <rank; I ++) 45 {46 for (int j = 0; j <rank2; j ++) 47 tempParptr [j] = parPtr [j]; 48 for (int j = 0; j <rank; j ++) 49 tempParptr [I + j * rank] = conPtr [j]; 50 tempDet = valDet (tempParptr, rank); 51 ansPtr [I] = tempDet/det; 52} 53} 54 void LinearEquations: showAns () 55 {56 if (ansPtr = NUL L) 57 {58 cout <"the linear equations have no or countless solutions" <endl; 59 return; 60} 61 cout <endl <"the solution of the linear equations is:" <endl; 62 for (int I = 0; I <rank; I ++) 63 cout <'X' <I + 1 <':' <ansPtr [I] <endl; 64} 65 int main () 66 {67 appDescribe (); 68 LinearEquations l1; 69 l1.sol (); 70 l1.showAns (); 71 system ("pause"); 72 return 0; 73} 74 void appDescribe () 75 {76 cout <"********************************* * ***** "<endl 77 <" this program uses the kramo law to solve the n-element linear equation. Group. "<Endl 78 <" case: "<endl 79 <" number of elements of the input equation: 4 "<endl 80 <" coefficient of input: "<endl 81 <" 1 1 1 1 "<endl 82 <" 1 2-1 4 "<endl 83 <" 2-3-1-5 "<endl 84 <" 3 1 2 11 "<endl 85 <" input constant: "<endl 86 <" 5-2-2 0 "<endl 87 <" the solution of the linear equations is: "<endl 88 <" x1: 1 "<endl 89 <" x2: 2 "<endl 90 <" x3: 3 "<endl 91 <" x4: -1 "<endl 92 <"****************************** * ******** "<endl; 93} 94 double valDet (double * detPtr, int rank) 95 {96 double val = 0; 97 if (rank = 1) return detPtr [0]; 98 for (int I = 0; I <rank; I ++) // The formula for calculating the remainder is stored in 99 {100 double * nextDetPtr = new double [(rank-1) * (rank-1)] In nextDetPtr []; 101 for (int j = 0; j <rank-1; j ++) 102 for (int k = 0; k <I; k ++) 103 nextDetPtr [j * (rank-1) + k] = detPtr [(j + 1) * rank + k]; 104 for (int j = 0; j <rank-1; j ++) 105 for (int k = I; k <rank-1; k ++) 106 nextDetPtr [j * (rank-1) + k] = detPtr [(j + 1) * rank + k + 1]; 107 val + = detPtr [I] * valDet (nextDetPtr, rank-1) * pow (-1.0, I); 108} 109 return val; 110} 111/* 112 data 2114 115 1 2116 2 1117 118 3 119 data 2: 120 4121 122 1 1 1123 1 2-1 4124 2-3-1-5125 3 1 2 11126 127 5-2 0128 */