题目
思路:
先判断是否为字母
再求和
通过迭代取余 计算0 1的数目
代码
alpha={'a':1,'b':2,'c':3,'d':4,'e':5,'f':6,'g':7,
'h':8,'i':9,'j':10,'k':11,'l':12,'m':13,'n':14,
'o':15,'p':16,'q':17,'r':18,'s':19,'t':20,'u':21,
'v':22,'w':23,'x':24,'y':25,'z':16}
input_=input()
sum=0
for i in input_:
if i =='\n':
break
if 'a'<=i<='z' or 'A'<=i<='Z':
i=i.lower()
sum=alpha[i]+sum
#q,r=divmod(sum,2)
zero_count = 0
one_count = 0
while(sum!=0):
sum, r = divmod(sum, 2)
if r==0:
zero_count=zero_count+1
else:
one_count = one_count +1
# if sum!=0 and r==0:
# zero_count=zero_count+1
# if sum != 0 and r == 1:
# one_count = one_count +1
print(zero_count,one_count)