function [x,N]= BSOR(A,b,x0,d,w,eps,M)
if nargin==5
    eps= 1.0e-6;
    M  = 10000;
elseif nargin<5
    error
    return
elseif nargin ==6
    M  = 10000;           %參數(shù)默認(rèn)值
end  

NS = size(A);
n = NS(1,1);
bnum = length(d);
bs = ones(bnum,1);
for i=1:(bnum-1)
    bs(i+1,1)=sum(d(1:i))+1;
    %獲得對(duì)角線上每個(gè)分塊矩陣元素索引的起始值
end

DB = zeros(n,n);       
for i=1:bnum
    endb = bs(i,1)+d(i,1)-1;
    DB(bs(i,1):endb,bs(i,1):endb)=A(bs(i,1):endb,bs(i,1):endb);
    %求A的對(duì)角矩陣
end
LB = -tril(A-DB);      %求A的下三角陣
UB = -triu(A-DB);      %求A的上三角陣

N = 0;
tol = 1;
iw = 1-w;
while tol>=eps 
    invDL = inv(DB-w*LB);
    x = invDL*(iw*DB+w*UB)*x0+w*invDL*b;   %塊迭代公式 
    N = N+1;
    tol = norm(x-x0);
    x0 = x;
    if(N>=M)
        disp('Warning: 迭代次數(shù)太多，可能不收斂！');
        return;
    end
end