雅可比迭代法

//节省时间不设置精度
//ps:注意收敛条件
#include <stdio.h>
#include <stdlib.h>
int n = 3;//3阶行列式 
double sum(int i,int a[3][3],double x0[3]){
    double result = 0.0;
    for(int j = 0 ; j < n ; j++){
        if( j != i){
            result += a[i][j] * x0[j];
        }
    }

    return  result;
}
int main(){

    int A[3][3] = {5, 1, 2, 3, 8, -4, -3, 2, 7};
    int b[3] = {21, -23, 15};
    //输出A和b 
    printf("A:\n");
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            printf("%d\t", A[i][j]);
        }
        printf("\n");
    }
    printf("b:\n");
    for(int i = 0; i < n; i++){
        printf("%d\t", b[i]);
    }
    printf("\n");
//  double e;
    double x0[3] = {0,0,0};//初始化向量 
    double t[3] = {0,0,0};//临时变量 
    double x[3] = {0,0,0};//k+1
    int times = 20;//迭代次数 
    for(int  k = 0; k < times; k++){
        for(int i = 0; i < n; i++){

        x[i] =  (b[i] - sum(i,A,x0)) / A[i][i] ;
        t[i] = x[i];    

        }
        printf("\n第%d次迭代:\n",k+1);
        for(int j = 0; j < n; j++){
            x0[j] = t[j];
            printf("x0[%d]=%.6lf\t",j,t[j]);
        }   //更新x0里的值并输出 
    }
    return 0;
}

测试结果：