Numeric Keypad

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题目描述

The numberic keypad on your mobile phone looks like below:

123

456

789

 0

 suppose you are holding your mobile phone with single hand. Your thumbpoints at digit 1. Each time you can 1)press the digit your thumbpointing at.2)moveyour thumb right,3)move your thumb down. Moving yourthumb left or up is not allowed.

 By using the numeric keypad under above constrains, you can producesome numbers like 177 or 480 while producing other numbers like 590 or52 is impossible.

 Given a number K, find out the maximum number less than or equal to Kthat can be produced.

输入描述:

the first line contains an integer T, the number of testcases.
Each testcase occupies a single line with an integer K.

For 50%of the data ,1<=K<=999.
For 100% of the data, 1<=K<=10^500,t<=20.

输出描述:

for each testcase output one line, the maximum number less than or equal to the corresponding K that can be produced.

输入例子:

3
25
83
131

输出例子:

25
80
129

代码演示：

# -*- encoding: utf8 -*-
"""
二维列表pad表示当前行值对应的数字下一次可以按的数字列在第二维度上
一维列表last表示当前数字的下一次按键可用数字数-1，方便循环进行了-1

举例解析：
数字：131
把1输入，
从3开始查找，有3种情况：
 1) 找到need，那么查找下一个字符
 2) 找不到need，那么试着去找第一个小于need的数，之后的数字就选择当前最大的合法值，结束并返回
 3) 连小于need的数都找不到，就像这个例子：在key=3的时候，找need=1，找不到，这个时候，
需要回到key=1的时候(去掉ret最后的字符，字符串变空需要特殊处理)，找1可用的数组里比3小的
数字(不能直接-1，因为可能不在可用范围内)。之后的数字填充最大的合法值，结束并返回。
"""

# mapping table
pad = [[0],
		   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
		   [0, 2, 3, 5, 6, 8, 9],
		   [3, 6, 9],
		   [0, 4, 5, 6, 7, 8, 9],
		   [0, 5, 6, 8, 9],
		   [6, 9],
		   [0, 7, 8, 9],
		   [0, 8, 9],
		   [9]]
last = [0, 9, 6, 2, 6, 4, 1, 3, 2, 0]

def find_min_number(num):
	"""
	find the minimum number
	"""
	ret = []
	input_str = str(num)
	input_length = len(input_str)
	ret.append(input_str[0])
	i = 1
	while i < input_length:
		need = int(input_str[i])
		key = int(ret[-1])
		j = last[key]
		# Situation 1)
		while j >= 0:
			if pad[key][j] == need:
				i += 1
				ret.append(str(pad[key][j]))
				break
			j -= 1 

		# Situation 2)
		if j < 0:
			j = last[key]
			while j >= 0:
				if pad[key][j] < need:
					ret.append(str(pad[key][j]))
					key = int(ret[-1])
					return ''.join(ret) + str(pad[key][last[key]])*(input_length-len(ret))
				j -= 1

		# Situation 3)
		if j < 0:
			need = key
			ret = ret[0:-1]
			if len(ret) == 0:
				ret.append(str(need-1))
				key = int(ret[-1])
				return ''.join(ret) + str(pad[key][last[key]])*(input_length-len(ret))

			key = int(ret[-1])
			j = last[key]
			while j >= 0:
				if pad[key][j] < need:
					ret.append(str(pad[key][j]))
					key = int(ret[-1])
					return ''.join(ret) + str(pad[key][last[key]])*(input_length-len(ret))
				j -= 1
	return ''.join(ret)

T = raw_input()
intT = int(T)

for i in range(intT):
	n = raw_input()
	num = int(n)
	print find_min_number(num)