Skip to main content

Merge Sort using Divide and Conquer in C

#include< stdio.h>
#include< conio.h>
#define MAX 50
void display(int[MAX],int);
void Merge_sort(int[MAX],int,int);
void Combine(int[MAX],int,int,int);
void main()
{
 int i,T[MAX],n;
 clrscr();


 printf("\n Enter number of element you want:");
 scanf("%d",&n);

 for(i=0;i < n;i++)
 {
  printf("\n Enter element[%d]:",i);
  scanf("%d",&T[i]);
 }
 printf("\n------------------------------------------");
 printf("\n Elements before sorting:");
 display(T,n);
 printf("\n------------------------------------------");

Merge_sort(T,0,n-1);

 printf("\n------------------------------------------");
 printf("\n Elements after sorting:");
 display(T,n);
 printf("\n------------------------------------------");

getch();
}
void Merge_sort(int T[MAX],int low,int high)
{
    int mid;
 if(high > low)
 {
  mid=(low+high)/2;
  Merge_sort(T,low,mid);
  Merge_sort(T,mid+1,high);
  Combine(T,low,mid,high);
 }

}
void Combine(int T[MAX],int low,int mid,int high)
{
 int i,j,k,temp[MAX];
 k=low;
 i=low;
 j=mid+1;
 
 while(i <= mid && j <= high)
 {
  if(T[i] <= T[j])
  {
   temp[k++]=T[i++];
  }
  else
  {
   temp[k++]=T[j++];
  }
 }
 while(i <= mid)
 {
  temp[k++]=T[i++];
 }
 while(j <= high)
 {
  temp[k++]=T[j++];
 }
 
 for(i=low;i <= high;i++)
  T[i]=temp[i];
}

void display(int T[MAX],int n)
{
 int i;
 for(i=0;i < n;i++)
 {
  printf(" %d ",T[i]);
 }
}

OUTPUT

 Enter number of element you want:7

 Enter element[0]:77

 Enter element[1]:55

 Enter element[2]:44

 Enter element[3]:66

 Enter element[4]:44

 Enter element[5]:22

 Enter element[6]:99

------------------------------------------------------------------------------------
 Elements before sorting: 77  55  44  66  44  22  99
------------------------------------------------------------------------------------
------------------------------------------------------------------------------------
 Elements after sorting: 22  44  44  55  66  77  99
------------------------------------------------------------------------------------

Comments

Popular posts from this blog

MVT (Multiprogramming Variable Task) in C Programming

#include #include void main() { int i,os_m,nPage,total,pg[25]; clrscr(); printf("\nEnter total memory size:"); scanf("%d",&total); printf("\nEnter memory for OS:"); scanf("%d",&os_m); printf("\nEnter no. of pages:"); scanf("%d",&nPage); for(i=0;i =pg[i]) { printf("\n Allocate page %d",i+1); total=total-pg[i]; } else printf("\n page %d is not allocated due to insufficient memory.",i+1); } printf("\n External Fragmentation is:%d",total); getch(); } OUTPUT Enter total memory size:1024 Enter memory for OS:256 Enter no. of pages:4 Enter size of page[1]:128 Enter size of page[2]:512 Enter size of page[3]:64 Enter size of page[4]:512 Allocate page 1 Allocate page 2 Allocate page 3 page 4 is not allocated due to insufficient memory. External Fragmentation is:64

First Come First Serve (FCFS) Page replacement algorithm in C Programming

#include #include int fsize; int frm[15]; void display(); void main() { int pg[100],nPage,i,j,pf=0,top=-1,temp,flag=0; clrscr(); printf("\n Enter frame size:"); scanf("%d",&fsize); printf("\n Enter number of pages:"); scanf("%d",&nPage); for(i=0;i OUTPUT Enter frame size:3 Enter number of pages:12 Enter page[1]:1 Enter page[2]:2 Enter page[3]:3 Enter page[4]:4 Enter page[5]:1 Enter page[6]:2 Enter page[7]:5 Enter page[8]:1 Enter page[9]:2 Enter page[10]:3 Enter page[11]:4 Enter page[12]:5 page | Frame content -------------------------------------- 1 | 1 -1 -1 2 | 1 2 -1 3 | 1 2 3 4 | 4 2 3 1 | 4 1 3 2 | 4 1 2 5 | 5 1 2 1 | 5 1 2 2 | 5 1 2 3 | 5 3 2 4 | 5 3 4 5 | 5 3 4 ---------------------------...

Deadlock Prevention using Banker’s Algorithm in C Programming

#include #include void main() { int allocated[15][15],max[15][15],need[15][15],avail[15],tres[15],work[15],flag[15]; int pno,rno,i,j,prc,count,t,total; count=0; clrscr(); printf("\n Enter number of process:"); scanf("%d",&pno); printf("\n Enter number of resources:"); scanf("%d",&rno); for(i=1;i OUTPUT Enter number of process:5 Enter number of resources:3 Enter total numbers of each resources:10 5 7 Enter Max resources for each process: for process 1:7 5 3 for process 2:3 2 2 for process 3:9 0 2 for process 4:2 2 2 for process 5:4 3 3 Enter allocated resources for each process: for process 1:0 1 0 for process 2:3 0 2 for process 3:3 0 2 for process 4:2 1 1 for process 5:0 0 2 available resources: 2 3 0 Allocated matrix Max need 0 1 0| 7 5 3| 7 4 3 3 0 2| 3 2 2| 0 2 0 3 0 2| 9 0 2| 6 0 0 2 1 1| ...