Reprint Please specify the source Thank you: http://blog.csdn.net/vmurder/article/details/42921985Title: give you a deck of cards, ask whether it is a draw (that is, add a card can "Hu"), if it is, the output to touch which cards can hu, not output "no"Hu rules: The first need to have two of the same number of cards are eliminated, and then can be three of the same card to eliminate, can also be three consecutive label cards to eliminate, eliminate the hu.For example, 22 444555 789 is the HU car
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Test instructions: Given 13 tiles, ask if you are "listening" to the card, if it is output "listen" which one.Analysis: This problem, very obvious violence, is on the original foundation put on a card, see is not can Hu, idea is very simple, also better realize, the result is tle, has been tle, this unscientific ah ...Very hard to write out, unexpectedly tle ... Heartache. is the first to determine a card, and then the flush analysis, in fact, is to prune, that is, if there are 1 or two cards, a
What about the Japanese mahjong .. There are no luxury seven pairs...
It's no more difficult than enumerative search.
Three types of cards
F1 is 7-to-F2 is 13-So F3 is a normal fake. First, find one and then find three.
Is always timeout .. Under the guidance of the peak, many pruning comments are marked .. In this way, it takes a long time ..
# Include
1. This question requires final answersCode(C #)
2. The final code must be compiled, run, and implement the following business functions
3. Time LimitOne hourIncluding the time for reading documents and submitting code
Business functions:
A number of mahjong cards are given (assuming there are only one million cards and no cards or cylinders)
The following conditions must be met:
All cards must be connected to any other card or three cards: 123
"88 Fan"
1 Senior XI is composed of 4 pairs of wind carving (staves) and cards. Regardless of the ring wind, Men, three wind carved, touch and
2 junior Yuan and card, there are 3 pairs of white engraved. Do not count the arrow engraved
3 green
Question link:
Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4431
Question:
Give a deck and find all the cards that can be pasted.
Solution:
Enumerate each card to see if it can be pasted.
Because there are only 14 cards in total, each time the
The original matrix and target matrix are given. Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4306
Two operations: select a horizontal line to exchange two elements (cost CA ).
Select the two-element exchange (CB) of a column ).
An operation
That is, 13 cards are given. Ask which cards can be added.
Hu Pai has the following situations:1. One pair + four groups of three identical cards or shunzi. Only M, S, and P can constitute a subfolder. Brands like East and West are not good.2. Seven
Plate:
East, south, west, north, middle, fa, white
Card:
Mei, lan, Zhu, chrysanthemum, spring, summer, autumn, and winter
Lettering:
Three identical
Shunzi:
Three in the same color form a sequence.
Card:
Four identical cards
Engraved child:
# Include
Int Hu (INT Pai [38]);Int remain (INT Pai [38]);
Int main (){// ° N percent» ¸ ± å æ å â ã µ? ~~×öà ''' â {}° {{{~~~~~~~~~{{??// ************** Pai [0], Pai [10], Pai [20], pai [30] ~~~~úö²» ó ~£%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
This is my last question in my career. Acm has left me too much... No.
The basic thing is to use enumeration + greedy.
But there are several minor issues:
1. Seven small pairs are seven different pairs.
2. east, west, north, south, and so on cannot
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