#include <stdio.h>
#include <stdlib.h>
#include <math.h>
/*
=====================================================================
Solving maths with the computers
=====================================================================

Ex 1) Brute force atack

Write a program in C language that can find (only) solutions like:

(82+8+1)^2
= 8281

(14+364+1)^2
= 143641

(5+1+2)^3
= 512

(6+72+4)^2
= 6724

(842+72+4)^2
= 842724

(62+726+4)^2
= 627264

(1+29+6)^2
= 1296

(929+29+6)^2
= 929296

(9+11+25)^3
= 91125

(5+715+36)^2
= 571536

(8+2+81)^2
= 8281

but omit cases such as as:

(9+12+25)^3
= 97336

(5+715+37)^2
= 573049

(8+21+81)^2
= 12100

The output of the program should be saved in txt file (the list should be extended by at least 10 more positive examples).

*/

#include <stdio.h>
#include <math.h>

//implementation of itoa (present in C++, but not in C)
void itoa(int n, char s[]) {
    int i = 0, sign = n;

    if (sign < 0) {
        n = -n;  // Handle negative numbers
    }

    do {
        s[i++] = n % 10 + '0';  // Extract digits by remainder
    } while ((n /= 10) > 0);

    if (sign < 0) {
        s[i++] = '-';
    }

    s[i] = '\0';

    // Reverse the string
    int start = 0, end = i - 1;
    while (start < end) {
        char temp = s[start];
        s[start] = s[end];
        s[end] = temp;
        start++;
        end--;

 }
}
int main() {
    FILE *fp;
    fp = fopen("/tmp/math_solutions.txt", "w");
    int max_A = 100;
    int max_B = 1000;
    int max_C = 1000;
    int max_power = 3;
    int max_solution_count = 10;
    int solution_counter = 0;

    for (int a = 1; a <= max_A && solution_counter<max_solution_count; ++a) {
        for (int b = 0; b <= max_B; ++b) {
            for (int c = 0; c <= max_C; ++c) {
                for (int d = 2; d <= max_power; ++d) {
                    int result = pow((a + b + c), d);

                    char abc_str[20];
                    //sprintf(abc_str, "%d%d%d", a, b, c);  //first option
                    //snprintf(abc_str, sizeof(abc_str), "%d%d%d", a, b, c); //second option

                      //third option, study it
                    char a_str[10];
                    char b_str[5];
                    char c_str[5];
                    itoa(a, a_str);
                    itoa(b, b_str);
                    itoa(c, c_str);
                    strcpy(abc_str, a_str);
                    strcat(abc_str, b_str);
                    strcat(abc_str, c_str);

                    int abc = atoi(abc_str);

                    if (result == abc) {
                        fprintf(fp, "(%d+%d+%d)^%d \n= %d \t(No. %d)\n\n", a, b, c, d, solution_counter+1, abc);
                        printf("(%d+%d+%d)^%d \t(No. %d) \n= %d\n\n", a, b, c, d, solution_counter+1, abc);
                        solution_counter++;
                    }
                }
            }
        }
    }

    fclose(fp);
    return 0;
}