Shortest Job First

#include <stdio.h>
#define N 12

struct process{
    char pid;
    int at;
    int et;
    int ct;
    int wt;
    int tt;
}pro[N] =   { //{'pid', at, et},
                {'A', 0, 10},
                {'B', 0, 3},
                {'C', 2, 2},
                {'D', 2, 1},
                {'E', 2, 4},
                {'F', 10, 5},
                {'G', 10, 7},
                {'H', 15, 8},
                {'I', 20, 10},
                {'J', 30, 9},
                {'K', 40, 3},
                {'L', 40, 2}
            };

int main(){
    int i, time = 0, sumWT = 0, sumTT = 0, sj = 0, done[N] = {0};

    while(1){
        for(i = 0; i < N; i++) if(!done[i] && time >= pro[i].at){sj = i; break;} //select first unexecuted process
        if(i == N) break; //Exit loop if all processes are done
        for(++i;i < N; i++){
            if(done[i] || time < pro[i].at) continue; //ignore unarrived & completed processes
            if(pro[i].et < pro[sj].et) sj = i; //if the execution time is better, make it sj
            else if((pro[i].et == pro[sj].et) && (pro[i].at < pro[sj].at)) sj = i; //if et is same, see at
            else if((pro[i].at == pro[sj].at) && (pro[i].pid < pro[sj].pid)) sj = i; //if at is same, see pid
        }
        if(time < pro[sj].at) {time++; continue;}//idle time
        done[sj] = 1;//sj has been executed
        pro[sj].ct = time + pro[sj].et;
        time = pro[sj].ct; //update time
        pro[sj].wt = pro[sj].ct - (pro[sj].at + pro[sj].et);
        pro[sj].tt = pro[sj].ct - pro[sj].at;
        sumWT += pro[sj].wt;
        sumTT += pro[sj].tt;
    }

    printf("\tpid\tAT\tET\tCT\tWT\tTT\n");
    for(i = 0; i < N; i++) printf("\t%c\t%d\t%d\t%d\t%d\t%d\n", pro[i].pid, pro[i].at, pro[i].et, pro[i].ct, pro[i].wt, pro[i].tt);
    printf("\nThe average waiting time is %.2f\nThe average turn around time is %.2f\n", sumWT/(float)N, sumTT/(float)N);
    return 0;
}