2021 Hefei informatics competition primary school group

1、 COVID-19 population immunity (covid)

COVID-19 has been in the world for nearly two years. , It has brought great trouble to all countries in the world , In order to defeat COVID-19 finally , Various countries are stepping up vaccine research and development , At present, our country has developed and issued inactivated vaccine 、 Adenovirus vector vaccine and other new coronavirus vaccines with different technologies , While meeting their own needs , It has also helped many countries , Together to curb the spread of the virus . Prevention is better than cure , Only by reaching a certain proportion of vaccination , That is to achieve mass immunization , That's the way to defeat COVID-19. . Studies have shown that , Different vaccine efficacy , The proportion of people who achieve mass immunization is different , Suppose that the vaccinated population of a country only reaches half of the total population of the country 75%, Group immunity can be achieved . Please calculate how many people need to be vaccinated in a country to achieve mass immunization .

Input :

The input data consists of 1 That's ok 1 A positive integer , Indicates the total population of a country , Unit: 10000 persons .

Output :

common 1 Line a positive integer , Indicates the number of people who need to be vaccinated to achieve mass immunization , rounding , Unit: 10000 persons .

Examples 1:

Input :(covid.in)

100

Output :(covid.out)

75

Data range :100≤ The total population ≤100000

2、 Organize books (book)

Another semester is over , I have accumulated many more books , You decided to tidy up , There are three operations in sorting , The following rules :

1 p It means to number p Put your book at the front

2 p It means to number p Put your books at the back

3 p q It means to number p Put your book under the number q Behind my book

1、2、3 Each represents the type of sorting operation ,p、q Indicates the number of the book , They are separated by spaces ; It is known that before sorting , All books from 1 Start numbering and discharging in sequence .

Input :

common m+1 That's ok . The first line has two positive integers separated by spaces n and m, respectively n This book

and m Finishing operation , Next m That's ok , Each row has 2 Or 3 A positive integer separated by a space ,

Corresponding to the above three sorting operations .

Output :

common 1 That's ok , The order of the sorted books , The books are separated by spaces .

Examples :

Input :(book.in)

10 4

1 3

2 4

3 3 6

3 1 5

Output :(book.out)

2 5 1 6 3 7 8 9 10 4

Data range :1≤n,m≤100000

3、 CY Craft (cycraft)
As the game goes on , You also get a lot of weapons and equipment , Every acquisition of a weapon and equipment requires a certain price . Suppose you have a timeline , It records that there is a weapon corresponding to a certain point in time , We give every point in time ti Weapons and equipment corresponding to the time i The power value of is wi. At this time, you have a time shuttle , You can shuttle through the time axis at will , Suppose you go back and forth to a point in time t, Definition t Time point to ti Point in time acquisition i The cost of weaponry is |t-ti|*wi. Please calculate which t The total cost of acquiring all weapons and equipment at a time point is the smallest , Output minimum total cost . Every weapon and equipment must be acquired from t Time to start , return t The cost of a point in time is 0;|t-ti| Express t-ti The absolute value of .
Input :

common n+1 That's ok , First line a positive integer n, Indicates the total number of weapons and equipment , Next n That's ok , Each row
Two positive integers separated by spaces , They represent time ti Corresponding weapons and equipment at that time
Power value wi.

Output :

common 1 Line an integer , Represents the minimum cost of acquiring all weapons .
Examples 1:
Input :(cycraft.in)
4
-1 7
0 2
7 3
3 4
Output :(cycraft.out)
40
Sample explanation : Shuttle to 0 The total cost of acquiring all weapons and equipment at any time :1*7+7*3+3*4=40. Wear
Shuttle to 7 The total cost of acquiring all weapons and equipment at any time :8*7+7*2+4*4=86 ...
Data range :1≤n≤10000,-1000≤ti,wi≤1000.

4、 Mice love food (mouse)

There are many people on one road 1 Start numbering the mouse's favorite food , Suppose there are an infinite number of delicious foods , And the mice flashed randomly next to any food , Then try some delicious food in turn , At least try the next food when it first appears , Ask which two adjacent delicacies are tried the most by rats , Output as many times as possible . Make sure each mouse appears and stops with a different food number .

Input :

The input data are n+1 That's ok , The first line shows the number of mice n, Next n Each row has two numbers ,

Respectively represent the food number of the mouse for the first time , And stop the food number

Output :

One positive integer per line , Indicates the maximum number of times .

Examples 1:

Input :(mouse.in)

3

1 4

2 5

3 7

Output :(mouse.out)

3

Sample explanation :

share 3 A mouse , The first one tried in turn 1-2-3-4 common 4 Grow food ; The second one tried in turn

2-3-4-5 Here you are 4 Chinese cuisine ; The third one tried in turn 3-4-5-6-7 common 5 Grow food . The adjacent 3-4

Food has been tried 3 Time .

Data range :2≤n≤10000, The variety of food is guaranteed to be in int Within the scope of , Each mouse appears and stops

The food numbers are different .