题解 | #合并表记录#
合并表记录
http://www.nowcoder.com/practice/de044e89123f4a7482bd2b214a685201
OMG...
This is a hard one which took me 2 hours to finish.
The use of sorted() makes dict {}
into list []
, which is worth noticing. It also turns the groups of keys and values into tuple - (key, value)
format.
d = {}
try:
l = int(input())
for i in range(0, l):
ipt = input().split(' ')
c_index = int(ipt[0])
c_value = int(ipt[1])
if c_index not in d.keys():
d[c_index] = c_value
else:
d[c_index] += c_value
out = sorted(d.items(), key = lambda x:x[0]) #items() is key and value!
# in lambda function, 0 means sort by key
j = len(out)
for k in range(0, j):
print(str(out[k][0]) + ' ' + str(out[k][1]))
except EOFError:
pass