function [word1,d]=cut22(d)
%寻找连续有文字的块，若长度大于某阈值，则认为该块有两个字符组成，需要分割
d=qiege(d);
[m,n]=size(d);
k1=1;
k2=1;
s=sum(d);
j=1;
while j~=n
    while s(j)==0
        j=j+1;
    end
    k1=j;
    while s(j)~=0 && j<=n-1
        j=j+1;
    end
    k2=j-1;
    if k2-k1>=round(n/6.5)
        [val,num]=min(sum(d(:,[k1+5:k2-5])));
        d(:,k1+num+5)=0;%分割
    end
end
%再切割
d=qiege(d);
%切割出7个字符
y1=10;
y2=0.25;
flag=0;
word1=[];
while flag==0
    [m,n]=size(d);
    left=1;
    wide=0;
    while sum(d(:,wide+1))~=0 %找到像素和为零的列
        wide=wide+1;
    end
    if wide<y1 %认为是左干扰 
        d(:,[1:wide])=0;%
        d=qiege(d);
    else
        temp=qiege(imcrop(d,[1 1 wide m]));
        [m,n]=size(temp);
        all=sum(sum(temp));
        two_thirds=sum(sum(temp([round(m/3):2*round(m/3)],:)));
        if two_thirds/all>y2
            flag=1;word1=temp;%word1
        end
        d(:,[1:wide])=0;d=qiege(d);
    end
end