1 subroutine zpassf5 (ido,l1,cc,ch,wa1,wa2,wa3,wa4)
2 implicit double precision (a-h,o-z)
4 1 wa1(1) ,wa2(1) ,wa3(1) ,wa4(1)
5 data tr11,ti11,tr12,ti12 /.309016994374947d0,-.951056516295154d0,
6 1-.809016994374947d0,-.587785252292473d0/
7 if (ido .ne. 2) go to 102
9 ti5 = cc(2,2,k)-cc(2,5,k)
10 ti2 = cc(2,2,k)+cc(2,5,k)
11 ti4 = cc(2,3,k)-cc(2,4,k)
12 ti3 = cc(2,3,k)+cc(2,4,k)
13 tr5 = cc(1,2,k)-cc(1,5,k)
14 tr2 = cc(1,2,k)+cc(1,5,k)
15 tr4 = cc(1,3,k)-cc(1,4,k)
16 tr3 = cc(1,3,k)+cc(1,4,k)
17 ch(1,k,1) = cc(1,1,k)+tr2+tr3
18 ch(2,k,1) = cc(2,1,k)+ti2+ti3
19 cr2 = cc(1,1,k)+tr11*tr2+tr12*tr3
20 ci2 = cc(2,1,k)+tr11*ti2+tr12*ti3
21 cr3 = cc(1,1,k)+tr12*tr2+tr11*tr3
22 ci3 = cc(2,1,k)+tr12*ti2+tr11*ti3
23 cr5 = ti11*tr5+ti12*tr4
24 ci5 = ti11*ti5+ti12*ti4
25 cr4 = ti12*tr5-ti11*tr4
26 ci4 = ti12*ti5-ti11*ti4
39 ti5 = cc(i,2,k)-cc(i,5,k)
40 ti2 = cc(i,2,k)+cc(i,5,k)
41 ti4 = cc(i,3,k)-cc(i,4,k)
42 ti3 = cc(i,3,k)+cc(i,4,k)
43 tr5 = cc(i-1,2,k)-cc(i-1,5,k)
44 tr2 = cc(i-1,2,k)+cc(i-1,5,k)
45 tr4 = cc(i-1,3,k)-cc(i-1,4,k)
46 tr3 = cc(i-1,3,k)+cc(i-1,4,k)
47 ch(i-1,k,1) = cc(i-1,1,k)+tr2+tr3
48 ch(i,k,1) = cc(i,1,k)+ti2+ti3
49 cr2 = cc(i-1,1,k)+tr11*tr2+tr12*tr3
50 ci2 = cc(i,1,k)+tr11*ti2+tr12*ti3
51 cr3 = cc(i-1,1,k)+tr12*tr2+tr11*tr3
52 ci3 = cc(i,1,k)+tr12*ti2+tr11*ti3
53 cr5 = ti11*tr5+ti12*tr4
54 ci5 = ti11*ti5+ti12*ti4
55 cr4 = ti12*tr5-ti11*tr4
56 ci4 = ti12*ti5-ti11*ti4
65 ch(i-1,k,2) = wa1(i-1)*dr2+wa1(i)*di2
66 ch(i,k,2) = wa1(i-1)*di2-wa1(i)*dr2
67 ch(i-1,k,3) = wa2(i-1)*dr3+wa2(i)*di3
68 ch(i,k,3) = wa2(i-1)*di3-wa2(i)*dr3
69 ch(i-1,k,4) = wa3(i-1)*dr4+wa3(i)*di4
70 ch(i,k,4) = wa3(i-1)*di4-wa3(i)*dr4
71 ch(i-1,k,5) = wa4(i-1)*dr5+wa4(i)*di5
72 ch(i,k,5) = wa4(i-1)*di5-wa4(i)*dr5