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
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