/** 
 * A brute force, single threaded find of "perfect" numbers.
 * It also finds minusone and plusone numbers 
 * (although I have never see it actually find a plusone).
 * 
 * A perfect number is one whose factors, excluding itself, 
 * add up to the number.  For example, 6.
 * 
 * Perfect numbers are rare.
 * https://en.wikipedia.org/wiki/List_of_perfect_numbers
 * 
 * @author gtowell
 * Created: April 2021
 * **/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>

typedef struct {
    int count;
    int data[1000];
} accumul;
/**
 * Print utility for the accumul struct
 * @param head -- a string to print
 * @param dat -- an instance of the structure
 * **/
void pres(char *head, accumul dat)
{
    for (int i = 0; i < dat.count; i++)
        printf("%s %2d %d\n", head, i, dat.data[i]);
}

accumul perfect;  // perfect numbers
accumul minusone; // number one less than perfect
accumul plusone; // numbers one more than perfect

long curr = 1; // the next number to check for perfection

/**
 * Check the next number for perfection
 * Uses the global curr rather than receiving the number as a param
 * If the number is perfect, store it in the global perfect structure
 * **/
void checkit() {
    int val = curr++;
    int j = sqrt(val);
    int sm = 1;
    for (int k = 2; k <= j; k++) {
        int div = val / k;
        if (k*div == val) {
            sm += k;
            if (k!=div)
                sm += div;
            }
        }
    //printf("%d   %d\n", i, sm);
    if (sm == val) {
        perfect.data[perfect.count++] = val;
    }
    if (sm==(val-1)) {
        minusone.data[minusone.count++] = val;
    }
    if (sm==(val+1)) {
        plusone.data[plusone.count++] = val;
    }
}

int main(int argc, char const *argv[])
{
    perfect.count = 0;
    plusone.count = 0;
    minusone.count = 0;

    curr = atol(argv[1]);
    long max = atol(argv[2]);
    while (1) {
        checkit();
        if (curr > max)
            break;
    }
    pres("perfect", perfect);
    pres("minusone", minusone);
    pres("plusone", plusone);
    printf("Complete %s %s\n", argv[1], argv[2]);
    return 0;
}