How to schedule multiple task or job on multiple processors?

int
main(int argc, char **argv)
{
    FILE *fp    = 0;
    int size    = 0;
    int cpus = 0;
    int iter    = 0;
    int *a      = 0;

    int *b      = 0;
    int *c      = 0;
    int bsize   = 0;
    int csize   = 0;
    int count   = 0;
    
    int sum     = 0;

    work *j     = 0;

    if(argc != 2)   {
        fprintf(stderr, "syntax:: r input_file\n");
        return 1;
    }

    fp = fopen(argv[1], "r");
    if(!fp) {
        fprintf(stderr, "error opening file [%s]\n", argv[1]);
        return 1;
    }
    // File format: size nr_cpus int1 int2 ... intsize
    fscanf(fp, "%d", &size);
    fscanf(fp, "%d", &cpus);
    if(!size || size < 0) {
        fprintf(stderr, "invalid size [%d]\n", size);
        fclose(fp);
        return 1;
    }
    if(!cpus || cpus < 0) {
        fprintf(stderr, "invalid cpus [%d]\n", cpus);
        fclose(fp);
        return 1;
    }
    fprintf(stdout, " size [%d] cpus [%d]\n", size, cpus);
    a  = (int *)calloc(1, size * sizeof(int));
    for(int i = 0; i < size; ++i) {
        fscanf(fp, "%d", &a[i]);
    }
    iter = _log2(cpus);
    if(iter == -1 || exp2(iter) != cpus) {
        fprintf(stderr, "invalid cpus not power of two cpus [%d] iter [%d]", cpus, iter);
        fclose(fp);
        return 1;
    }
    fprintf(stdout, " log2 [%d]\n", iter);
    work_q .push(new work(a, size, 0));
    for(int i = 0; i < iter; ++i) {
        j = work_q.front();
        if(!j) {
            break;
        }
        while(j && j->pass == i) {
            work_q.pop();
            p(j->a,j->size,&b,&bsize,&c,&csize);
            if(bsize)
                work_q.push(new work(b,bsize,j->pass+1));
            if(csize)
                work_q.push(new work(c,csize,j->pass+1));
            free((void *)j->a);
            delete j;
            j = work_q.front();
        }
    }
    // The final schedule is in the queue.
    fprintf(stdout, " queue size [%d]\n", work_q.size());
    size = work_q.size();
    for(count = 0; count < size; ++count) {
        printf("cpu = [%d]\n",count);
        j = work_q.front();
        sum = 0;
        printf(" len = [%d] { ", j->size);
        for(int k = 0; k < j->size; ++k) {
            printf("%d ", j->a[k]);
            sum += j->a[k];
        }
        printf("} \n");
        printf(" sum = [%d]\n", sum);
        free((void *)j->a);
        delete j;
        work_q.pop();
    }
    fclose(fp);
    return 0;
}