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Catalog

  One 、 Sum the greatest common divisor and the least common multiple

Two 、 Hollow square pattern

  3、 ... and 、 Arrow pattern

  Four 、 Civil service interview

  One 、 Sum the greatest common divisor and the least common multiple

Problem description ：

Xiao Lele recently learned in class how to find the maximum common divisor and minimum common multiple of two positive integers , But he can't find the sum of the maximum common divisor and the minimum common multiple of two positive integers , Please help him solve the problem .

Input description ：

Each set of inputs contains two positive integers n and m.(1 ≤ n ≤ 109,1 ≤ m ≤ 109) Output description

Output description ：

For each group of input , Output a positive integer , by n and m The sum of the greatest common divisor and the least common multiple of .

Example ：

Input ：

10 20

Output ：

30

* Problem analysis ：

What we need to know here is the least common multiple =（m*n）/ greatest common divisor ;

Code implementation ：

#include<stdio.h>
int main()
{
   long long m;
   long long n;
   scanf("%d %d",&m,&n);
    long long p= m*n;
    while(n!=0)
    {
       long long t=m%n;
        m=n;
        n=t;
    }
    long long max=m;
    long long min=p/m;
    printf("%lld",max+min);
}

Two 、 Hollow square pattern

Problem description ：

KiKi I learned the cycle ,BoBo The teacher gave him a series of printing exercises , The task is to print “*” Composed of “ Hollow ” Square pattern .

Input description ：

Multi group input , An integer （3~20）, Represents the number of rows output , It also means that the sides of a square are made up of “*” The number of .

Output description ：

Enter... For each line , For output “*” Composed of “ Hollow ” Square , Every “*” There is a space after .

Example ：

Input ：

4

Output ：

* * * * * * * * * * * * 

️️ Problem analysis ：

We can see that this question uses * Output a square surrounded by * And the middle is a pattern of spaces , Then we can define a two-dimensional array , Then print by cycling , Control judgment conditions , Assign a value of * still " ".

Code implementation ：

#include<stdio.h>
void print(int a)
{
    int i,j;
    for(i=0;i<a;i++)
    {
        for(j=0;j<a;j++)
        {
            if(i==0 || i==a-1 || j  == 0  ||  j == a-1)
            { 
                printf("* ");
            }
            else
                printf("  ");
        }
        printf("\n");
    }
}
int main()
{
    int a=0;
    while(scanf("%d",&a)!=EOF)
    {
        print(a); 
    }
    return 0;
}

Running results ：

  3、 ... and 、 Arrow pattern

Input description ：

Multiple groups of input , One integer per row （2~20）.

Output description ：

Enter... For each line , For output “*” The shape of an arrow .

Example 1：

Input ：

2

Output ：

 * ** *** ** *

Example 2：

Input ：

3

Output ：

 * ** *** **** *** ** *

Problem analysis ：

We also need to find the right method for this problem , We can find that this is a symmetrical figure up and down , So we can try to print the graphics on the top half first , Then print the graphics on the bottom half in the same way , utilize for Cycle control printing , See code comments for specific public functions .

Code implementation ：

#include <stdio.h>
int main()
{
	int a;
	while(scanf("%d",&a)!=EOF)// Meet the cycle requirements 
	{
		for(int i=1;i<=a+1;i++)// According to the requirements of the topic, first show the upper half of the arrow 
	    {
	    	for(int j=1;j<=(a+1-i)*2+i;j++)// Combine loops and judgments to determine spaces and * The number of 
	    	{
	    		if(j<=(a+1-i)*2)
	    		{
	    			printf(" ");
				}
				else
				{
					printf("*");
				}
			}
			printf("\n");// Pay attention to the position of line breaks ;
		}
		for(int m=1;m<=a;m++)// Show the lower half of the arrow in the same way 
		{
			for(int n=1;n<=m*2+a+1-m;n++)
			{
				if(n<=(m*2))
				{
					printf(" ");
				}
				else
				{
					printf("*");
				}
			}
			printf("\n");
		 } 
	}
	return 0;
}

Running results ：

  Four 、 Civil service interview

Problem description ：

Civil servant interview on-site scoring . Yes 7 Examiner , Enter several groups of grades from the keyboard , Each group 7 A score （ Percentage system ）, Take out the highest score and the lowest score , Output the average score of each group .

（ notes ： There are many groups of input ）

Input description ：

Every line , Input 7 It's an integer （0~100）, representative 7 Achievements , Separate... With spaces .

Output description ：

Every line , Output the average score excluding the highest and lowest scores , After the decimal point 2 position , Wrap after output of each line .

Example 1：

Input ：

99 45 78 67 72 88 60

Output ：

73.00

️️ Problem analysis ：

This question is relatively simple , We design sort The function first finds out the highest score and the lowest score , Then calculate the seven scores and subtract the highest and lowest scores , Finally, find the average value , Pay attention to our count = 0; max = 0;small = 100; sum = 0; These four parts need to be reset to the initial value after our judgment ！🧂🧂

Code implementation ：

#include <stdio.h>
int main()
{
	int a, max = 0, small = 100, sum = 0, count = 0;
	while (scanf("%d", &a) != EOF)
	{
		if (a > max)// Determine the highest score 
		{
			max = a;
		}
		if (a < small)// Determine the lowest score 
		{
			small = a;
		}
		sum += a;
		count++;// Counter 
		if (count == 7)// Counter =7 When the score represents a group, it can be calculated 
		{
			printf("%.2f\n", (sum - max - small) / 5.0);
			count = 0;// Reset 
			max = 0;// Reset 
			small = 100;// Reset 
			sum = 0;// Reset 
		}
		
	}

	return 0;
}

Running results ：

  Okay , That's all for today's problem analysis , Thank you for reading !