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Reverse output three digit and arithmetic sequence
2022-07-07 23:39:00 【Yuesi】
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;
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