Short Job First(sjf) based on arrival time(non-preemptive)

/* Simulate the following cpu scheduling algorithms.             '
    B. Short Job First(sjf) based on arrival time(non-preemptive).    */

#include<stdio.h>
#include<conio.h>
void swap(int *x,int *y)
{
    int *temp;
    *temp=*x;
    *x=*y;
    *y=*temp;
}
void main()
{
    int i,j,bt[10],bt1[10],n,twt=0,temp1,z[20],p[10];
    int wt[10],sum=0,sum1=0,at[10],k,l;
    float avgt=0.00;
    clrscr();
    printf("\n Enter no. of Processes ::");
    scanf("%d",&n);
    printf("\n Enter the %d burst times::",n);
    for(i=0;i<n;i++)
    {
        printf("\n\n Enter burst and arrival times for process%d::",i+1);
        scanf("%d%d",&bt[i],&at[i]);
        p[i]=i+1;
    }
    i=j=0;
    while(at[j+1]<bt[i])
        j++;

    for(k=0;k<j;k++)
        bt1[k]=bt[k+1];
    /*for(i=0;i<j;i++)
        printf("%6d",bt1[i]);*/
    for(i=1;i<j-1;i++)
    {
        for(k=i+1;k<j;k++)
        {
            if(bt[i]>bt[k])
                swap(&bt[i],&bt[k]);
        }
    }
/*    for(i=0;i<j;i++)
        printf("\n%4d",bt1[i]); */
    for(i=0;i<n;i++)
    {
        if(i==0)
        {
            wt[i]=0;
            sum=sum+bt[i];
            sum1=sum1+bt[i];
        }
        else
        {
            sum=sum+bt[i];
            wt[i]=sum-at[i];

        }
        sum1=sum1+bt[i];
        z[i]=sum1;

    }
    /*for(i=0;i<n;i++)
        printf("%d",p[i]);   */
    printf("\n\n----------------------------------------------\n");
    printf("\n\tPNo\tbt\tat\twt\n");
    printf("\n----------------------------------------------\n");
    for(i=0;i<n;i++)
    {
        twt+=wt[i];
        printf("\n\n\tP%d\t%d\t%d\t%d\n",p[i],bt[i],at[i],wt[i]);
    }
    printf("\n----------------------------------------------\n");
    avgt=(float)twt/(float)n;
    printf("\n\nGannt Chart ::\n\n");
    printf("\t0");
    for(i=1;i<n;i++)
        printf("%4d",z[i]);
    printf("\n\n Total waiting time is ::%d",twt);
    printf("\n\n Average waiting time is ::%f",avgt);
    getch();
}

/* Input and Output :-


 Enter no. of Processes ::3
 Enter the 3 burst times::
 Enter time for process1::24
 Enter time for process2::3
 Enter time for process3::3
  231
----------------------------------------------
    PNo     bt      wt
----------------------------------------------
    P2      3       0
    P3      3       3
    P1      24      6
----------------------------------------------

Gannt Chart ::

    0   3   6

 Total waiting time is ::9
 Average waiting time is ::3.000000        */