fortran实现jacobi

下面是Fortran实现Jacobi方法的代码：

program jacobi

implicit none

integer :: n, i, j, k, itermax, iter
real :: tol, diff, sum
real, dimension(100,100) :: a, b, x, xnew

! 读入矩阵维数和迭代次数
print *, "Enter the matrix size:"
read *, n
print *, "Enter the maximum number of iterations:"
read *, itermax

! 读入矩阵元素和右端向量
print *, "Enter the matrix elements:"
do i = 1, n
    do j = 1, n
        read *, a(i,j)
    end do
end do
print *, "Enter the right-hand side vector:"
do i = 1, n
    read *, b(i)
end do

! 初始化解向量
x = 0.0
xnew = 0.0

! 设置迭代停止条件
tol = 1.0e-6

! 开始迭代
do iter = 1, itermax
    ! 计算新的解向量
    do i = 1, n
        sum = b(i)
        do j = 1, n
            if (i /= j) then
                sum = sum - a(i,j)*x(j)
            end if
        end do
        xnew(i) = sum/a(i,i)
    end do

    ! 计算当前解向量和新的解向量之间的差异
    diff = 0.0
    do i = 1, n
        diff = diff + abs(xnew(i) - x(i))
    end do

    ! 更新解向量
    x = xnew

    ! 判断是否满足停止条件
    if (diff < tol) then
        print *, "Convergence achieved after ", iter, " iterations."
        exit
    end if
end do

! 如果迭代次数达到了最大值，仍然没有满足停止条件，则认为算法失败
print *, "Iteration failed to converge within ", itermax, " iterations."

end program jacobi