Niuke real problem programming - day15

Compile environment ：c++

1、ZJ3 Programming questions 2

describe

One contains only ’a’ and ’b’ Two character string s, The length is n, Each operation can convert a character once （ Put one ’a’ Set to ’b’, Or put one ’b’ Into a ’a’); But there is an upper limit on the number of operations m, Ask within a limited range of operands , What is the length of the substring that can get the largest continuous same character .

Algorithmic thought ：

Use the idea of sliding window , First, traverse the string , First, get a subsequence that contains at most （m individual a/m individual b）, That is, let the subsequence a perhaps b The number of operations is just critical m when . Then slide the window , Only when satisfied again a Sum of numbers b The number of is greater than m, And one of the characters just exceeds m The condition of , Compare the subsequence length again . After the traversal is completed, the number of operations that meet , The longest subsequence .

The code part implements ：

2、 Fishing competition

describe ：

ss please cc Go fishing at home , Fish ponds can be divided into n＊m Lattice of , Each grid has different probability of catching fish ,cc Always in coordinates (x,y) Grid fishing , and ss Catch a grid randomly every minute . ask t Who of them has a high probability of catching at least one fish in minutes ？ How much? ？

Algorithmic thought ：

The problem is actually a basic math problem , According to the probability problem ,t The probability of everyone catching fish in minutes can be marked as P event , Yes P=1-(1-P（ss/cc）)^t, And for Pcc event , It can be seen that the probability of catching fish per minute is the coordinate element value , and Pss Randomly select the fishing port , Then his probability of catching per minute is the average of the sum of the probabilities of all grids .

The code part implements ：

3、 The annual toy game

describe ：

At the annual meeting of a company , The organizer arranged a small game to adjust the atmosphere . The rules of the game are as follows ：

N Individual participation in the game , Stand in a row to rob the workers M A little doll . In order to increase the fun and difficulty of the game , The rules stipulate , People participating in the game can't grab more than two or more gifts than people on both sides , Otherwise, you will be punished . At the end of the game, the person who has the most dolls will win a grand prize .

Suppose everyone wants to win this prize , Please stand at No K The trick of three positions is to win the game , At most, you can have several dolls ？

Algorithmic thought ：

To make the first k The little trick of bit wins the game without violating the rules , Then the number of dolls of small tricks and adjacent participants can only decrease in turn 1 individual . Let the trick get the most dolls first , Then traverse the participants on the left and right side in descending order , Add the number of dolls of everyone in turn . At first, the total number of dolls must be greater than the actual number of dolls , Then let the number of dolls of the trick decrease in turn , When the number of dolls allocated in a cycle does not exceed the actual number of dolls , This is the optimal allocation scheme , Can make small moves under the rules , Get the most dolls . Just print it out , Pay attention to the data processing of abnormal input .

The code part implements ：