Beginner level Luogu 1 [sequence structure] problem list solution

Catalog

Super Marie

Letter conversion

  Number reversal

Regreasing soap water

The swimming time of small fish

Primary school mathematics N close 1

Triangle area

Apples and worms

Diagonals

Late for school

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【 introduction 1】 Sequential structure - List of questions - Luogu | New ecology of computer science education (luogu.com.cn)

Super Marie

• C++ For multiline output R"( )"

#include<iostream>
using namespace std;
int main() {
cout<<R"(                ********
               ************
               ####....#.
             #..###.....##....
             ###.......######              ###            ###
                ...........               #...#          #...#
               ##*#######                 #.#.#          #.#.#
            ####*******######             #.#.#          #.#.#
           ...#***.****.*###....          #...#          #...#
           ....**********##.....           ###            ###
           ....****    *****....
             ####        ####
           ######        ######
##############################################################
#...#......#.##...#......#.##...#......#.##------------------#
###########################################------------------#
#..#....#....##..#....#....##..#....#....#####################
##########################################    #----------#
#.....#......##.....#......##.....#......#    #----------#
##########################################    #----------#
#.#..#....#..##.#..#....#..##.#..#....#..#    #----------#
##########################################    ############)"<<endl;
	return 0;
}

Letter conversion

• subject ： Enter a lowercase letter , Output its corresponding capital letter . For example, the input q[ enter ] when , Will be output Q.
• Be careful ：toupper The return is int value , To return a letter , Plus char, namely char(toupper())

  #include<iostream>
  #include<ctype.h>
  using namespace std;
  int main(){
  char a;
  cin>>a;
  cout<<char(toupper(a));
  return 0;
  }

  

  Number reversal

•               utilize scanf and printf

#include<iostream>
using namespace std;
int main(){
char a,b,c,d;
scanf("%c%c%c.%c",&a,&b,&c,&d);
printf("%c.%c%c%c",d,c,b,a);
return 0;
}

Regreasing soap water

• Enter three decimal places %.3f

printf("%.3f\n",t/n);

The swimming time of small fish

 

#include<iostream>
using namespace std;
int main(){
int a,b,c,d;
cin>>a>>b>>c>>d;
// On the same day ,c Must be greater than a, Just judge d and b
if(d>=b)
{
cout<<c-a<<" "<<d-b;  
}
else
{
cout<<c-a-1<<" "<<d+60-b;   
}
return 0;
}

Primary school mathematics N close 1

• Be careful 1.0/4 and 1/4 The difference between
• If it is 1/4 The result is 0, 1/3 Empathy however 1.0/4 Will come to 0.25
• 

 1/4 int type , 1.0/4 double type ;  1/4.0 double type It is also concluded that 0.25

The result of the expression is the result of the highest type  

else if(T==13)
{cout<<(int)(pow(1.0*4/3*pi*1064,1.0/3))<<endl;}

Triangle area

• sqrt Square root function

 

Apples and worms

• There are a few intact apples left , Then a few mouthfuls of apple must not count , The final result is int Directly erase the decimal point

#include<iostream>
using namespace std;
int main(){
double m,t,s;
cin>>m>>t>>s;
if(t==0)  // Be careful when eating an apple 0 minutes , It means eating apples quickly , In the end, there must be 0 An apple .
{cout<<"0"<<endl;}
else if((int)(m-s/t)<=0){ 
cout<<"0"<<endl;}
else{
cout<<(int)(m-s/t)<<endl;}
return 0;
}

Diagonals

• Four vertices have a diagonal intersection  
• unsigned long long Range ratio long long Large scope  
• unsigned long long The range is [0,2^64-1].long long The range is [-2^63-1,2^63+1]

• n and n-1 There must be one ２ Multiple , therefore ２ It can be divided into ; Empathy n,n-1,n-2 One of them must be ３ Multiple , therefore ３ It can be divided into （ Get rid of ２ It will only eliminate the factor ２ And yes ３ No impact ); Empathy ４ You can also get rid of

#include<iostream>
using namespace std;
int main(){
unsigned long long n;
cin>>n;
cout<<n * (n-1) / 2 * (n-2) / 3 * (n-3) / 4<<endl;
return 0;
}

 

Late for school

• ceil（） Rounding up

#include<iostream>
#include<math.h>
using namespace std;
int main() {

	double s, v;
	int hour = 7, min;
	cin >> s >> v;
	int total = ceil(s / v) + 10;  //ceil()  Rounding up  3/2=2  ceil(s/v) eg： cost 3.2 minute , according to 4 Minute calculation 
	int totalhour = total / 60;
	min = 60 - total % 60;  //70min,  6:50  60-10
	if (total % 60 == 0)
		min = 00;
	while (totalhour--)
	{
		hour--; // If only write if (hour < 0)  When hour = 24; when    There are 24：30 hour from 7 Began to set , The same goes for the day before ,
		if (hour < 0 && min != 00)
			hour = 23;
		else if (hour < 0)
			hour = 24;
	}
	
	printf("%02d:%02d",hour ,min );
	return 0;
}