[Solved] Non-Preemptive priority scheduling

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The correct code is:

#include <stdlib.h>
#include <stdio.h>


void main()
{

    int pn = 0;                 //Processes Number
    int CPU = 0;            //CPU Current time
    int allTime = 0;        // Time neded to finish all processes
    printf("Enrer Processes Count: ");
    scanf("%d",&pn);
    int AT[pn];
    int ATt[pn];
    int NoP = pn;
    int PT[pn];             //Processes Time
    int PP[pn];             //Processes piriorty
    int PPt[pn];
    int waittingTime[pn];
    int turnaroundTime[pn];
    int i=0;
    //Scanning Time and Piriorty
    for(i=0 ;i<pn ;i++){
        printf("\nProcessing time for P%d: ",i+1);
        scanf("%d",&PT[i]);
        printf("Piriorty for P%d: ",i+1);
        scanf("%d",&PP[i]);
        PPt[i] = PP[i];
        printf("Arrival Time for P%d: ",i+1);
        scanf("%d",&AT[i]);
        ATt[i] = AT[i];
    }




    int LAT = 0;        //LastArrivalTime
    for(i = 0; i < pn; i++)
        if(AT[i] > LAT)
            LAT = AT[i];

    int MAX_P = 0;        //Max Piriorty
    for(i = 0; i < pn; i++)
        if(PPt[i] > MAX_P)
            MAX_P = PPt[i];




    int ATi = 0;        //Pointing to Arrival Time indix
    int P1 = PPt[0];     //Pointing to 1st piriorty Value
    int P2 = PPt[0];     //Pointing to 2nd piriorty Value


    //findding the First Arrival Time and Highst piriorty Process
    int j = -1;
    while(NoP > 0 && CPU <= 1000){
        for(i = 0; i < pn; i++){
            if((ATt[i] <= CPU) && (ATt[i] != (LAT+10))){
                if(PPt[i] != (MAX_P+1)){
                    P2 = PPt[i];
                    j= 1;

                    if(P2 < P1){
                        j= 1;
                        ATi = i;
                        P1 = PPt[i];
                        P2 = PPt[i];
                    }
                }
            }
        }

        if(j == -1){
            CPU = CPU+1;
            continue;
        }else{


            waittingTime[ATi] = CPU - ATt[ATi];
            CPU = CPU + PT[ATi];
            turnaroundTime[ATi] = CPU - ATt[ATi];
            ATt[ATi] = LAT +10;
            j = -1;
            PPt[ATi] = MAX_P + 1;
            ATi = 0;        //Pointing to Arrival Time indix
            P1 = MAX_P+1;     //Pointing to 1st piriorty Value
            P2 = MAX_P+1;     //Pointing to 2nd piriorty Value
            NoP = NoP - 1;

        }




    }



    printf("\nPN\tPT\tPP\tAT\tWT\tTT\n\n");
    for(i = 0; i < pn; i++){
       printf("P%d\t%d\t%d\t%d\t%d\t%d\n",i+1,PT[i],PP[i],AT[i],waittingTime[i],turnaroundTime[i]);
    }

    int AvgWT = 0;
    int AVGTaT = 0;
    for(i = 0; i < pn; i++){
        AvgWT = waittingTime[i] + AvgWT;
        AVGTaT = turnaroundTime[i] + AVGTaT;
    }


   printf("AvgWaittingTime = %d\nAvgTurnaroundTime = %d\n",AvgWT/pn,AVGTaT/pn);

}



/*
Test Cases:
PT: Processing Time
PP: Process priority
WT Waitting Time
TaT: Turnaround Time
Arrival time for 1st 2 cases is 0

PN      PT      PP      WT      TaT

P1      10      3       6       16
P2      1       1       0       1
P3      2       4       16      18
P4      1       5       18      19
P5      5       2       1       6


PN      PT      PP      WT      TaT

P1      1       1       0       1
P2      2       2       1       3
P3      3       3       3       6
P4      4       4       6       10
P5      5       5       10      15



PN      PT      PP      AT      WT      TT

P1      3       2       0       0       3
P2      5       6       2       11      16
P3      4       3       1       2       6
P4      2       5       4       7       9
P5      9       7       6       12      21
P6      4       4       5       2       6
P7      10      10      7       20      30



PN      PT      PP      AT      WT      TT

P1      4       2       0       0       4
P2      3       3       1       3       6
P3      1       4       2       5       6
P4      5       5       3       5       10
P5      2       5       4       9       11

*/```

solved Non-Preemptive priority scheduling