Reverse output three digit and arithmetic sequence

Markdown Answer key
Count garlic customers to output three digits in reverse ：
Original link
Calculate the last term of the garlic guest equal difference series ：
Original link

subject

Garlic has a three digit number , She wants you to output the three digits in reverse .

Input format

A three digit number n

Output format

Reverse output n, To keep the leading 0.
Extra space at the end of each line when outputting , It doesn't affect the correctness of the answer

Data range

100≤n≤999

The sample input

100

Sample output

001

The question

Read in a three digit integer , Output it from right to left

Ideas

• 1. The integer n Read in as a whole , Create three integer variables to represent their bits, tens and hundreds , Output in the order of one bit, ten bits and hundreds
• 2. The integer n Divide by digits 3 Integers are read in sequence and stored in three integer variables , Output in the reverse order of read in storage
• 3. take n Read in as a whole （ example ： hypothesis n by 123）, establish 3 Variables are stored in turn n A bit of （a=3） ten (b=2) Hundred bit (c=1), And take the bits as the required output （ And n It's reverse , Such as 100 And 001） Top 100 , Ten as ten , Hundreds as a bit （ example ： With n=123 For example Output number =3100+210+1*1=321 ）
  Pay attention to whether there is 0！！！
• 4. The capacity of the created array is greater than or equal to 3, From the subscript for 0 Start storing into the array , And subscript from 2 Start to 0 Traverse the array and output the corresponding value （ available for loop ）

Pit point

• 1. When the number of reverse output Output as a whole When , Pay attention to whether there is 0, Such as n=100 The reverse output is not 001 It is 1.（ The output is not 3 digit ）
  Solution
  Output to form three digits

      scanf("%03d", Variable name ); ( This variable represents the relation with n Inverse integer )

Code

Will work with n The reverse number is output three times according to the number of digits

#include<stdio.h> 
int main(){
    
	int n; 
	scanf("%d",&n);
	int a1,a2,a3;
	a1=n%10;
	//a1=n%100 It's OK ;  Single digit  
	a2=n/10%10;
	// Ten digits  
	//405 except 10 be equal to 40 more than 5 405/10=40 40 except 10 be equal to 4 more than 0 
	//40%10=0 
	a3=n/100;
	// Hundreds of digits   example 405 except 100 be equal to 4 more than 5
	// 405/100=4 405%100=5 
	printf("%d%d%d",a1,a2,a3);
	return 0;
}

Will work with n The reverse number is output as an integer at one time

#include<stdio.h>
int main(){
    
	int n;
	int gewei,shiwei,baiwei;
	int sum=0;// The assignment is 0 Don't miss  
	scanf("%d",&n);
	gewei=n%10;//gewei=n%100; It's OK 
	shiwei=n/10%10;
	baiwei=n/100;
	sum=gewei*100+shiwei*10+baiwei;
	// And n reverse , But not necessarily in three digits 
	printf("%03d",sum); 
	return 0;
}

summary

1. Cultivate the good habit of looking at the range of input and output data

2. Give full consideration to special circumstances

…

subject

The arithmetic sequence is a very interesting sequence , The difference between any two adjacent terms of it is equal .

Garlic gentleman gives the first two terms of an arithmetic sequence a1,a2, Please n What is the item? .

Input format

a line , Contains three integers a1, a2, n.

Output format

An integer , That is to say n Item value .

Data range

-100 ≤ a1, a2 ≤ 100, 0<n≤1000.

The sample input

1 4 100

Sample output

298

The question

The first term and the second term of the known arithmetic sequence are known , Output No n term

Ideas

• 1. Find the tolerance of the arithmetic sequence d, utilize an=a1+(n-1)*d Formula evaluation
• 2. Use the relationships of the series to evaluate ,ak+as=aj+al;( Conditions k+s=j+l)
• 3. according to n Range build array （ The capacity is greater than 1000, You can also wait , But not recommended ）, Find the value of each term
  Output the value of the corresponding subscript in the array as required

Pit point

• 1.an The expression formula of must be remembered , Otherwise, we can only find the value by pushing the law

Code

#include<stdio.h> 
int main(){
    
	int a1=0,a2=0,n=0;// The first one is , The second item , And the number of bits to output  
	int d;// Save tolerance d Equal to the last term of the sequence minus the previous term  d=a2-a1=a3-a2=a4-a3......=an-an-1 
	scanf("%d",&a1);
	scanf("%d%d",&a2,&n);
	d=a2-a1;
	printf("%d\n",a1+(n-1)*d); 
	//an=a1+(n-1)*d;
	// an It's the... Of the arithmetic sequence n term  
	//a1 It's the first one  d Is the tolerance equal to the last item of the sequence minus the previous item  
	return 0;
}

summary

The formula of accumulating the sequence of equal differences ,an=a1+(n-1)*d
When d If you don't know, you can subtract any two terms to get
Such as ：ak-as=(k-s)*d;