C language programming training topics: K characters in left-handed string, little Lele and Euclidean, printing arrow pattern, civil servant interview, poplar matrix

Catalog

One 、 In left-handed string K Characters .

Two 、 Little Lele and Euclid （ Find the greatest common divisor ）

3、 ... and 、 Print arrow pattern

Four 、 Civil service interview

5、 ... and 、 Young's matrix

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One 、 In left-handed string K Characters .

Title Description ：ABCDE Turn a character left to get BCDEA, Left handed two get CDEAB

How easy it is ：

#include<stdio.h>

int my_strlen(char* str)// Find the number of characters 
{
	int count = 0;// Record the number of characters 
	while (*str != '\0')
	{
		count++;
		str++;
	}
	return count;
}

void left_rotate(char arr[], int k)
{
	int i = 0;
	int len = my_strlen(arr);
	
	for (i = 0; i < k; i++)
	{
		// Left hand character steps 
	    //1. Save the characters to be rotated first 
		char tmp = arr[0];

		//2. Move the remaining characters forward 
		int j = 0;
		for (j = 0; j < len - 1; j++)
		{
			arr[j] = arr[j + 1];
		}

	    //3. Store the previously stored characters into the empty bits of the array 
		arr[len - 1] = tmp;
	}
	
}

int main()
{
	int k = 0;
	scanf("%d", &k);
	char arr[] = "ABCDEF";
	left_rotate(arr, k);
	printf(" The result of left rotation ：%s\n", arr);
	return 0;
}

Two 、 Little Lele and Euclid （ Find the greatest common divisor ）

Title 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 .

How easy it is ：

#include<stdio.h>


int main()
{
	long long a = 0;
	long long b = 0; 
	long long comax = 0;
	long long comin = 0;
	long long k = 0;
	scanf("%lld %lld", &a, &b);
	k = a * b;
	while (a && b)
	{
		if (a > b)
		{
			a = a % b;
		}
		else
		{
			b = b % a;
		}
	}
	comax = a > b ? a : b;
	comin = k / comax;
	printf("%lld\n", comax + comin);
}

3、 ... and 、 Print arrow pattern

Title Description ： Utilization cycle . Use * Print arrow pattern

How easy it is ：

#include<stdio.h>

int main()
{
	int n, i, j;
	while (scanf("%d", &n) != EOF)
	{
		for (i = 0; i < n + 1; i++) // Control the top half * Output （n+1 That's ok ）
		{
			for (j = i; j < n; j++)  // Control space output 
			{
				printf("  ");
			}
			for (j = 0; j <= i; j++)  // control * Output 
				
			{
				printf("*");
			}
			printf("\n");  // Line break 
		}
		for (i = 0; i < n; i++)  // Control the lower half * Output （n That's ok ）
		{
			for (j = 0; j <= i; j++)
			{
				printf("  ");
			}
			for (j = i; j < n; j++)
			{
				printf("*");
			}
			printf("\n");
		}
	}
	return 0;
}

Four 、 Civil service interview

Title 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 ）

How easy it is ：

#include <stdio.h>

int main()
{
	int a = 0;
    int max = 0;// Maximum 
    int small = 100;// minimum value 
    int sum = 0;
    int 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;
}

5、 ... and 、 Young's matrix

Title Description ：

There is a matrix of numbers , Each row of the matrix is incremented from left to right , The matrix is increasing from top to bottom , Please write a program to find out whether a number exists in such a matrix .

requirement ： Time complexity is less than O(N);

How easy it is ：

#include<stdio.h>

find_k(int arr[3][3], int k, int* row, int* col)
{
	int x = 0;// That's ok 
	int y = *col - 1;// Column 
	while (x <= *row && y >= 0)
	{
		// The number to be found is less than the number in the upper right corner of the matrix 
		if (k < arr[x][y])
		{
			// Matrix column number minus 1
			y--;
		}
		else if (k > arr[x][y])
		{
			// Matrix rows minus 1
			x++;
		}
		else
		{
			// The number to be found is equal to the number in the upper right corner of the matrix  -  eureka 
			*row = x;
			*col = y;
			return 1;// eureka 
		}
	}
	*row = -1;
	*col = -1;
	return -1;// Did not find 
}

int main()
{
	int arr[3][3] = { 1,2,3,4,5,6,7,8,9 };
	int k = 0;// The number to find 
	scanf("%d", &k);
	int row = 3;// That's ok 
	int col = 3;// Column 
	int ret = find_k(arr, k, &row, &col);
	if (1 == ret)
	{
		printf(" eureka , The subscript is ：%d %d", row, col);
	}
	else
	{
		printf(" Can not find \n");
	}
	return 0;
}