# Chapter I preliminary programming

## a master hand 's first small display

1. Output a line of "Hello World" on the screen and wrap it.

```#include <iostream>
using namespace std;
int main() {
cout << "Hello World" << endl;
return 0;
}
```

2. Calculation questions - 1 and 1.0
Calculation 1 + 1 / (1 + 1 / (1 + 1 / 5))

```#Method 1
#include <iostream>
using namespace std;
int main() {
cout<< fixed << setprecision(4) << 1.0+1.0/(1.0+1.0/(1.0+1.0/5))<<endl;
// fixed and setprecision (...) are format controllers,
// Fixed means to output floating-point numbers in fixed-point format (other formats include scientific counting method)
// Setprecision (4) means that 4 digits are reserved after the decimal point when outputting floating-point numbers
return 0;
}
#Method 2
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
printf("%.4f",1.0+1.0/(1.0+1.0/(1.0+1.0/5)));
return 0;
}
```

3. Calculation problem - trigonometric function
[Title Description]
Calculate the value of y in the following formula

```#include <iostream>
#include<cmath>
using namespace std;
int main() {
cout<< fixed << setprecision(5) << pow(sin(3.14159/4),2) + sin(3.14159/4)*cos(3.14159/4) - pow(cos(3.14159/4),2)<<endl;
// fixed and setprecision (...) are format controllers,
// Fixed means to output floating-point numbers in fixed-point format (other formats include scientific counting method)
// Setprecision (4) means that 4 digits are reserved after the decimal point when outputting floating-point numbers
return 0;
}
#Method 2:
#include<cstdio>
#include<cmath>
int main() {
printf("%.5f",pow(sin(3.14159/4),2) + sin(3.14159/4)*cos(3.14159/4) - pow(cos(3.14159/4),2));
return 0;
}
```

# Chapter 2 variables and algebraic thinking

## a master hand 's first small display

1. Basic operation
[Title Description]
Xiao Ming drew a round cake. The radius of this cake is 2.52384. How big is the area of this cake? (pi = 3.1415926)

```#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
float r=2.52384;
float pi=3.1415926;
float s;
cout<< fixed << setprecision(4) << pi*r*r <<endl;
return 0;
}
```

2. Acceleration
[Title Description]
Small I performs uniform acceleration. The initial velocity is 0 m/s. After t seconds, the velocity is v m/s. what is the acceleration (m/s^2)?
[enter description]
Enter two integers V, t. (0 < v, t <= 1000)
[output description]
Output the acceleration of the object with 4 decimal places.

3. Find the area of the triangle
[Title Description]
Given the length of the three sides of a triangle (meeting the composition conditions of the triangle), calculate the area of the triangle.
[enter description]
The input has only one line, which is three real numbers a, B, c (a > 0, b > 0, c > 0), representing the side lengths of the three sides of the triangle.
[output description]
The output has only one line, which is a real number S (accurate to two decimal places), representing the area of the triangle.

```#include <iostream>
#include <iomanip>
#include<cmath>
using namespace std;
int main()
{
float a,b,c;
cin >>a>>b>>c;
float p=(a+b+c)/2;
float s=sqrt(p*(p-a)*(p-b)*(p-c));
cout<< fixed << setprecision(2) << s <<endl;
return 0;
}
```

# Logical reasoning and enumeration problem solving

## a master hand 's first small display

1. [Title Description]
Six suspects were arrested in the baswick murder. Their confessions are as follows:
A: I'm not a criminal
B: One of A and C is A criminal
C: A and B lied
D: C and F lied
E: Of the other five, only A and D told the truth
F: I'm a criminal
Only half of them told the truth and there was only one murderer.
The answer to this question is not unique. Please program to find out all possible killers.
[enter description]
nothing
[output description]
Output all answers in alphabetical order. Each line is represented as an answer.

```#include <iostream>
using namespace std;
int main() {
for (int criminal = 1; criminal <= 6; criminal++)//loop
{
int count = 0;
if (criminal!= 1)
count++;
if (criminal == 1||criminal == 3)
count++;
if (criminal != 6)
count++;
if (criminal != 1 && criminal != 3 && criminal != 6)
count++;
if (criminal == 6)
count++;
if (count == 3)
{
if (criminal == 1)
cout << "A" << endl;
else if (criminal == 2)
cout << "B" << endl;
else if (criminal == 3)
cout << "C" << endl;
else if (criminal == 4)
cout << "D" << endl;
else if (criminal == 5)
cout << "E" << endl;
else if (criminal == 6)
cout << "F" << endl;
}
}
return 0;
}
```

2. Print triangles
[Title Description]
Please program a solid isosceles right triangle with output length n.
[enter description]
Enter an integer n. (0 < n <= 10)
[output description]
Outputs a solid isosceles right triangle of length n.

# There is no problem in running, but there is a problem in submitting. Do you know the reason? The code is as follows?

```#include <iostream>
using namespace std;
int main()
{
int i=0,j=0,n;
cin>>n;
for (i=0;i<=n;i++)
{
for (j=n-i+1;j<=n;j++)
{
cout<<"*";
}
cout<<endl;
}
return 0;
}
```

3. Judge leap year
[Title Description]
Enter a year to determine whether it is a leap year.
[enter description]
Enter an integer a. (1000 <= a <= 10000)
[output description]
If it is a leap year, output "YES", otherwise output "NO".

```#include <iostream>
using namespace std;
int main()
{
int year;
cin >> year;
if (((year%400)==0)||(((year%100)!=0))&&((year%4)==0))
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
return 0;
}
```

# Chapter IV screening method and search

## a master hand 's first small display

1. Joseph Ring
[Title Description]
Joseph problem (sometimes called Josephus permutation) is a problem that appears in computer science and mathematics. In computer programming algorithms, similar problems are also called Joseph rings.

3. Quality factor decomposition
[Title Description]
Decompose the prime factors of n positive integers greater than 1, sort them from small to large, and output them in equation format.
[enter description]
The first line is an integer n, which represents the number of prime factors to be decomposed. (1<=n<=100)
Lines 2 to n+1 are n positive integers x that need to be decomposed_ i. (1 < x_i < 10^9)
[output description]
There are n lines in total. Each line decomposes the equation obtained by the prime factor, and the factors are arranged from small to large.

Posted on Mon, 29 Nov 2021 06:40:26 -0500 by MerlinJR