问三角化码的算法复杂度

EN

y=0;
// n times
for k=1:n
    // some constant complexity stuff
    // it basically sets y to floor(k/2)
    if(mod(k,2)==0)
        y=y+1;
    end
    // n-k times
    for j=k+1:n
        A(k,j)=A(k,j)-tau*U(k,k);
        // executed everytime except the last few iteration of the outermost loop
        if (k<=n-2)
            // constant complexity
            for i=n-1:n
                A(i,j)=A(i,j)-tau*U(i,k);
            end
        end
        // executes y-2 times
        for i=n+1:n-2+y
            // so we basically need to bound the number of times this is executed
            A(i,j)=A(i,j)-tau*U(i,k);
        end            
    end
end