Highest Response Ratio Next

#include <stdio.h>
#define N 9
struct process{
    char pid;
    int at;
    int et;
    float pri;
    int rem;
    int done;
    int ct;
    int wt;
    int tt;
}pro[N] =   { //{'pid', at, et},
                {'A', 0, 10},
                {'B', 0, 5},
                {'C', 0, 2},
                {'D', 0, 4},
                {'E', 10, 8},
                {'F', 10, 16},
                {'G', 10, 6},
                {'H', 20, 3},
                {'I', 20, 1}
            };
void init(){ //initially, all processes are unfinished and their remaining and execution times are same.
    int i;
    for(i=0; i< N; i++){
        pro[i].done=0;
        pro[i].rem=pro[i].et;
    }
}
int nextProc(int time){
    int i, send=0;
    for(i = 0; i < N; i++)
        if(pro[i].done==0) break; //find first process that's not completed.
    if(i==N) return -1;
    send=i; //send will be returned at the end of function
    for(; i < N; i++){
        if(pro[i].done || pro[i].at > time) continue; //if the process is completed, skip it
        if(time==0||time==10||time==20)
                pro[i].pri=((float)(time-pro[i].at)/(float)pro[i].rem)+1; //calculated at t=0,10,20...
        if (pro[i].pri > pro[send].pri) send = i; //process with highest response time will be returned
        else if(pro[i].pri == pro[send].pri && pro[i].et < pro[send].et) send = i; //compare execution time
        else if(pro[i].et == pro[send].et && pro[i].at < pro[send].at) send = i; //compare arrival times
        else if(pro[i].at == pro[send].at && pro[i].pid < pro[send].pid) send = i; //compare pids
    }
    if(time==0||time==10||time==20){
        printf("\n\nAt t=%d, the processes already arrived but not completed are:-\n", time);
            for(i=0; i < N; i++)
                if(pro[i].done || pro[i].at > time) continue;
                else printf("%c(Pri: %.3f)",pro[i].pid, pro[i].pri);
        printf("\n\n");
    }
    return send;
}
void printResult(int sumWT, int sumTT){
    int i;
    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);
}
int main(){
    int i, time = 0, sumWT = 0, sumTT = 0,  curr;
    init();
    while(1){
        curr = nextProc(time); //process to be executed
        if(curr == -1) break; //if all processes have completed, print the result
        printf("At time=%d, %c gets CPU.(RR: %.3f, remaining: %d)\n", time, pro[curr].pid, pro[curr].pri, pro[curr].rem);
        pro[curr].rem--; //subtract one time unit from the process's remaining time
        time++;
        if(pro[curr].rem == 0){
            pro[curr].done=1; //if remaining time is 0, the process is completed.
            pro[curr].ct = time; //completed time is the time when process execution is completed
            pro[curr].tt = pro[curr].ct - pro[curr].at;
            pro[curr].wt = pro[curr].tt - pro[curr].et;
            printf("\n%c has finished executing at %d\n\n", pro[curr].pid, time);
            sumWT += pro[curr].wt;
            sumTT += pro[curr].tt;
        }
    }
    printResult(sumWT, sumTT);
    return 0;
}