Assignment 6. Q1. Create a database of students using structures, where in each entry of the database will have the following fields:

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1 Assignment 6 Q1. Create a database of students using structures, where in each entry of the database will have the following fields: 1. a name, which is a string with at most 128 characters 2. their marks in physics which is an int between 0 and their marks in chemistry which is an int number between 0 and their marks in mathematics which is an int number between 0 and 100 You have to output a list of students in the following order.. 1. if a student 'A' has lower marks in physics than a student 'B', then A's data is listed before B. 2. If A and B have the same physics marks and A has lower chemistry marks than B, then A is listed before B. 3. If A and B have the same marks in physics and chemistry, and A has lower marks in mathematics than B, then A is listed before B. 4. If all marks are equal and A's name precedes B's name in the dictionary order, then A is listed before B. Input Format : First line contains the number of students n, where 1<=n<=100. In following n lines each line contains(space separated) a name and their respective marks in physics, chemistry, maths, where 0<=marks<=100. Output Format : Sorted database of n lines. Sol. #include<stdio.h> #include<stdlib.h> struct student char name[20]; int phy,che,math; ; typedef struct student student;

2 void print(student *s,int n) int i; for(i=0;i<n;i++) printf("%s\t",s[i].name); printf("%d\t",s[i].phy); printf("%d\t",s[i].che); printf("%d\n",s[i].math); int comparator(const void *p, const void *q) float l = ((student *)p)->phy; float r = ((student *)q)->phy; if((l-r)!=0) return (l-r); else l = ((student *)p)->che; r = ((student *)q)->che; if((l-r)!=0) return(l-r); else l = ((student *)p)->math; r = ((student *)q)->math; return (l-r); int main() int i,n; scanf("%d",&n); student *student_info= (student *)malloc(sizeof(student)*n);

3 for(i=0;i<n;i++) scanf("%s",student_info[i].name); scanf("%d",&student_info[i].phy); scanf("%d",&student_info[i].che); scanf("%d",&student_info[i].math); qsort((void *)student_info,n,sizeof(student),comparator); print(student_info,n); return 0; Q.2 A graph is abstractly a collection of vertices which are labelled by non-negative integers, and a collection of edges. A graph called an undirected graph if we talk of merely the presence of an edge between vertices i and j, rather than its direction. For example, the following is a graph: In this problem, you are given the edges in an undirected graph. An edge is a pair of non-negative integers (i, j) which indicates that the vetex i is connected to vetex j by an edge. Afterwards, you will be given a vertex number n. You have to output the list of vertices which are connected n by an edge, in the order in which the edges were input. Input You are given the following. 1. The first line contains an integer, E, between 1 and This is followed by E lines, where each containing a pair of numbers i and j where i and j are both non-negative integers <= 34,000. No edge will be listed more than once. 3. The last line contains a non-negative integer n <= 34,000. n is assured to be a vertex listed in one of the E lines in part (2). Output You have to output the list of nodes to which n has an edge, in the order in which the edges were input, one line for each vertex.

4 Sol #include <stdio.h> #include <stdlib.h> struct node int vertex; struct node *next; ; struct list_entry struct node *head; struct node *tail; ; struct list_entry list_entries[34000]; void init_list_entries() int i; for ( i=0 ; i<100 ; i++ ) list_entries[i].head = list_entries[i].tail = NULL; struct node * make_node ( int data ) struct node *temp = (struct node *)malloc(sizeof(struct node)); temp -> vertex = data; temp -> next = NULL; return temp;

5 void insert_at_end(int a, int b) struct node *node1; struct node *node2; node1 = make_node (a); if(list_entries[b].head == NULL) list_entries[b].head = node1; list_entries[b].tail = node1; else list_entries[b].tail->next = node1; list_entries[b].tail = node1; node2 = make_node(b); if(list_entries[a].head == NULL) list_entries[a].head = node2; list_entries[a].tail = node2; else list_entries[a].tail->next = node2; list_entries[a].tail = node2; return; void print_adjacent_vertices_of(int n)

6 struct node *current = list_entries[n].head; while(current!= NULL) printf("%d\n",current->vertex); current = current->next; return; int main() int num_edges; int a; int b; int n; int i=0; scanf("%d", &num_edges); for ( i=0; i<num_edges ; i++) scanf ( "%d", & a); scanf ( "%d", & b); insert_at_end(a,b); scanf("%d",&n); print_adjacent_vertices_of(n); Q.3 You are given a sequence of integers terminated with a -1. The -1 is not part of the input sequence.

7 Next, you are given a positive number N. You have to create a linked list with the input sequence of integers as entries. You can use the following structure. struct node int data; struct node *next; ; Now, you have to delete all but the last N elements from the linked list, and print the resulting list. (i.e. The resulting list will consist of only the last N elements from the list.) If N is longer than the length of the linked list, you must print -1. While printing, the entries of the list must be separated by a single space. Sample Input Sample Output You are given a sequence of integers terminated with a -1. The -1 is not part of the input sequence. Next, you are given a positive number N. You have to create a linked list with the input sequence of integers as entries. You can use the following structure. struct node int data; struct node *next; ; Now, you have to delete all but the last N elements from the linked list, and print the resulting list. (i.e. The resulting list will consist of only the last N elements from the list.)

8 If N is longer than the length of the linked list, you must print -1. While printing, the entries of the list must be separated by a single space. Sample Input Sample Output Sol #include <stdio.h> #include <stdlib.h> struct node int data; struct node *next; ; int list_length; struct node* create_node ( int n ) struct node *new_node; new_node = (struct node *) malloc ( sizeof(struct node) ); new_node->data = n; new_node->next = NULL; return new_node; struct node* create_list ( struct node *head ) int number;

9 struct node *current_node, *new_node; scanf("%d",&number); if(number!= -1) head = create_node ( number ); current_node = head; list_length++; scanf ( "%d", &number ); while ( number!= -1 ) new_node = create_node ( number ); current_node->next = new_node; list_length++; current_node = new_node; /* advance to next node */ scanf ( "%d", &number ); return head; /* Delete the first num nodes from the list */ struct node *delete_first ( int num, struct node *head ) int i=1; struct node *current_node = head; struct node *next_node; while ( i <= num ) next_node = current_node->next; free(current_node);

10 current_node = next_node; i++; head = current_node; return head; void print_list ( struct node *head ) struct node * current_node; if ( head == NULL ) return; current_node = head; while ( current_node!= NULL ) printf ( "%d ", current_node->data ); current_node = current_node->next; printf("\n"); return; int main() int number; int pruned_length=0; struct node *head;

11 head = create_list ( head ); scanf("%d", &pruned_length); if(pruned_length > list_length) printf ( "-1\n" ); return 0; else head=delete_first(list_length-pruned_length, head); print_list(head); return 0;