Dados los registros de los estudiantes con cada registro que contiene la identificación, el nombre y la edad de un estudiante. Escriba un programa en C para leer estos registros y mostrarlos ordenados por edad o id.
Clasificación por edad
Ejemplos:
Input: Student Records = {
{Id = 1, Name = bd, Age = 12 },
{Id = 2, Name = ba, Age = 10 },
{Id = 3, Name = bc, Age = 8 },
{Id = 4, Name = aaz, Age = 9 },
{Id = 5, Name = az, Age = 10 } }
Output:
{{Id = 3, Name = bc, Age = 8 },
{Id = 4, Name = aaz, Age = 9 },
{Id = 2, Name = ba, Age = 10 },
{Id = 5, Name = az, Age = 10 },
{Id = 1, Name = bd, Age = 12 } }
Planteamiento: Este problema se resuelve en los siguientes pasos:
- Cree una estructura con los campos id, nombre y edad.
- Leer los registros de los estudiantes en la estructura.
- Defina un comparador configurando reglas para la comparación. Aquí la edad se puede ordenar con la ayuda de la diferencia de la edad de 2 estudiantes. (Estudiante1 -> edad – Alumno2 -> edad)
- Ahora ordene la estructura según el comparador definido con la ayuda del método qsort() .
- Imprime los expedientes de los alumnos ordenados.
Programa:
// C program to read Student records
// like id, name and age,
// and display them in sorted order by Age
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// struct person with 3 fields
struct Student {
char* name;
int id;
char age;
};
// setting up rules for comparison
// to sort the students based on age
int comparator(const void* p, const void* q)
{
return (((struct Student*)p)->age - ((struct Student*)q)->age);
}
// Driver program
int main()
{
int i = 0, n = 5;
struct Student arr[n];
// Get the students data
arr[0].id = 1;
arr[0].name = "bd";
arr[0].age = 12;
arr[1].id = 2;
arr[1].name = "ba";
arr[1].age = 10;
arr[2].id = 3;
arr[2].name = "bc";
arr[2].age = 8;
arr[3].id = 4;
arr[3].name = "aaz";
arr[3].age = 9;
arr[4].id = 5;
arr[4].name = "az";
arr[4].age = 10;
// Print the Unsorted Structure
printf("Unsorted Student Records:\n");
for (i = 0; i < n; i++) {
printf("Id = %d, Name = %s, Age = %d \n",
arr[i].id, arr[i].name, arr[i].age);
}
// Sort the structure
// based on the specified comparator
qsort(arr, n, sizeof(struct Student), comparator);
// Print the Sorted Structure
printf("\n\nStudent Records sorted by Age:\n");
for (i = 0; i < n; i++) {
printf("Id = %d, Name = %s, Age = %d \n",
arr[i].id, arr[i].name, arr[i].age);
}
return 0;
}
Producción:
Unsorted Student Records: Id = 1, Name = bd, Age = 12 Id = 2, Name = ba, Age = 10 Id = 3, Name = bc, Age = 8 Id = 4, Name = aaz, Age = 9 Id = 5, Name = az, Age = 10 Student Records sorted by Age: Id = 3, Name = bc, Age = 8 Id = 4, Name = aaz, Age = 9 Id = 2, Name = ba, Age = 10 Id = 5, Name = az, Age = 10 Id = 1, Name = bd, Age = 12
Clasificación por identificación
Ejemplos:
Input: Student Records = {
{Id = 1, Name = bd, Age = 12 },
{Id = 2, Name = ba, Age = 10 },
{Id = 3, Name = bc, Age = 8 },
{Id = 4, Name = aaz, Age = 9 },
{Id = 5, Name = az, Age = 10 } }
Output:
{{Id = 1, Name = bd, Age = 12 },
{Id = 2, Name = ba, Age = 10 },
{Id = 3, Name = bc, Age = 8 },
{Id = 4, Name = aaz, Age = 9 },
{Id = 5, Name = az, Age = 10 } }
Planteamiento: Este problema se resuelve en los siguientes pasos:
- Cree una estructura con los campos id, nombre y edad.
- Leer los registros de los estudiantes en la estructura.
- Defina un comparador configurando reglas para la comparación. Aquí la identificación se puede ordenar con la ayuda de la diferencia de la identificación de 2 estudiantes. (Estudiante1 -> id – Estudiante2 -> id)
- Ahora ordene la estructura según el comparador definido con la ayuda del método qsort() .
- Imprime los expedientes de los alumnos ordenados.
Programa:
// C program to read Student records
// like id, name and age,
// and display them in sorted order by ID
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// struct person with 3 fields
struct Student {
char* name;
int id;
char age;
};
// setting up rules for comparison
// to sort the students based on ID
int comparator(const void* p, const void* q)
{
return (((struct Student*)p)->id - ((struct Student*)q)->id);
}
// Driver program
int main()
{
int i = 0, n = 5;
struct Student arr[n];
// Get the students data
arr[0].id = 1;
arr[0].name = "bd";
arr[0].age = 12;
arr[1].id = 2;
arr[1].name = "ba";
arr[1].age = 10;
arr[2].id = 3;
arr[2].name = "bc";
arr[2].age = 8;
arr[3].id = 4;
arr[3].name = "aaz";
arr[3].age = 9;
arr[4].id = 5;
arr[4].name = "az";
arr[4].age = 10;
// Print the Unsorted Structure
printf("Unsorted Student Records:\n");
for (i = 0; i < n; i++) {
printf("Id = %d, Name = %s, Age = %d \n",
arr[i].id, arr[i].name, arr[i].age);
}
// Sort the structure
// based on the specified comparator
qsort(arr, n, sizeof(struct Student), comparator);
// Print the Sorted Structure
printf("\n\nStudent Records sorted by ID:\n");
for (i = 0; i < n; i++) {
printf("Id = %d, Name = %s, Age = %d \n",
arr[i].id, arr[i].name, arr[i].age);
}
return 0;
}
Producción:
Unsorted Student Records: Id = 1, Name = bd, Age = 12 Id = 2, Name = ba, Age = 10 Id = 3, Name = bc, Age = 8 Id = 4, Name = aaz, Age = 9 Id = 5, Name = az, Age = 10 Student Records sorted by ID: Id = 1, Name = bd, Age = 12 Id = 2, Name = ba, Age = 10 Id = 3, Name = bc, Age = 8 Id = 4, Name = aaz, Age = 9 Id = 5, Name = az, Age = 10