本文主要是介绍【数值计算方法】雅可比解线性方程，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
废话少说，直接上干货。
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MaxSize 100
double A[MaxSize][MaxSize]; //系数矩阵
double B[MaxSize]; //系数矩阵
double C[MaxSize][MaxSize]; //去对角线矩阵
double D[MaxSize][MaxSize]; //储存D逆
double E[MaxSize][MaxSize]; //储存-(D-1)*(L+U)
double F[MaxSize]; //(D-1)*B
double X[MaxSize];
double X1[MaxSize];
double Y[MaxSize];
#define e 1e-6 //10的负6次方
int n; //矩阵尺寸大小// 初始化矩阵
void InitMatrix()
{int i, j;for (i = 0; i < n; i++){for (j = 0; j < n; j++){if (i == j)D[i][j] = A[i][j];elseC[i][j] = A[i][j];}}
}// 雅可比迭代求解线性方程组
int Jacobi()
{int k;double sum, max_diff;for (k = 0; ; k++) {max_diff = 0.0;for (int i = 0; i < n; i++){sum = 0.0;for (int j = 0; j < n; j++){if (j != i)sum += A[i][j] * X[j];}X1[i] = (B[i] - sum) / A[i][i];double diff = fabs(X1[i] - X[i]);if (diff > max_diff)max_diff = diff;}for (int i = 0; i < n; i++)X[i] = X1[i];if (max_diff < e) // 满足精度要求退出循环break;}return k; // 返回迭代次数
}// 输入系数矩阵和向量
void input()
{int i, j;printf("请输入系数矩阵A：\n");for (i = 0; i < n; i++)for (j = 0; j < n; j++)scanf_s("%lf", &A[i][j]);printf("请输入向量B：\n");for (i = 0; i < n; i++)scanf_s("%lf", &B[i]);printf("请输入初始向量X：\n");for (i = 0; i < n; i++)scanf_s("%lf", &X[i]);
}// 打印方程组的近似解
void print()
{int i;printf("方程组的近似解：\n");for (i = 0; i < n; i++)printf("%lf\n", X[i]);
}int main()
{printf("请输入矩阵行数：\n");scanf_s("%d", &n);printf("\n");input();InitMatrix();int N =Jacobi();print();printf("迭代次数为：%d",N);return 0;
}
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