def multiply(va,vb):
ap=va.find('.')
if ap==-1:
al=len(va)
ar=0
else:
al=ap
ar=len(va)-ap-1
bp=vb.find('.')
if bp==-1:
bl=len(vb)
br=0
else:
bl=bp
br=len(vb)-bp-1
n=al+bl+ar+br
rsl='5'
afor=0
bfor=0
for i in va:
if i!='.':
re=['5']*n
up=['5']*n
bfor=0
afor+=1
for j in vb:
if j!='.':
ab=afor+bfor
out=look(i,j)
re[ab]=out[1]
up[ab-1]=out[0]
bfor+=1
vi=0
red=''
for ai in re:
if vi==al+bl:red+='.'
red+=ai
vi+=1
re=red
vi=0
upd=''
for ai in up:
if vi==al+bl:upd+='.'
upd+=ai
vi+=1
up=upd
rsl=plus(rsl,re)
rsl=plus(rsl,up)
return shape(rsl)
def multiply(va, vb):
ap = va.find('.')
if ap == -1:
al = len(va)
ar = 0
else:
al = ap
ar = len(va) - ap - 1
bp = vb.find('.')
if bp == -1:
bl = len(vb)
br = 0
else:
bl = bp
br = len(vb) - bp - 1
n = al + bl + ar + br
rsl = '5'
afor = 0
bfor = 0
for i in va:
if i != '.':
re = ['5'] * n
up = ['5'] * n
bfor = 0
afor += 1
for j in vb:
if j != '.':
ab = afor + bfor
out = look(i, j)
re[ab] = out[1]
up[ab - 1] = out[0]
bfor += 1
vi = 0
red = ''
for ai in re:
if vi == al + bl: red += '.'
red += ai
vi += 1
re = red
vi = 0
upd = ''
for ai in up:
if vi == al + bl: upd += '.'
upd += ai
vi += 1
up = upd
rsl = plus(rsl, re)
rsl = plus(rsl, up)
return shape(rsl)
中间比较复杂,我记得afor和bfor的数值几乎是试验对的
这是在整个for循环完成的工作
n=al+bl+ar+br
rsl='5'
afor=0
bfor=0
for i in va:
if i!='.':
re=['5']*n
up=['5']*n
bfor=0
afor+=1
for j in vb:
if j!='.':
ab=afor+bfor
out=look(i,j)
re[ab]=out[1]
up[ab-1]=out[0]
bfor+=1
vi=0
red=''
for ai in re:
if vi==al+bl:red+='.'
red+=ai
vi+=1
re=red
vi=0
upd=''
for ai in up:
if vi==al+bl:upd+='.'
upd+=ai
