multiplication.c~

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/* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
 * Description: Finds all multiplications of the form                          *
 *                            *  *  *                                          *
 *                        x      *  *                                          *
 *                            -------                                          *
 *                         *  *  *  *                                          *
 *                      *  *  *  *                                             *
 *                   ----------------                                          *
 *                      *  *  *  *  *                                          *
 * in such a way that every star is replaced with a digit                      *
 * that is a prime number (2, 3, 5 or 7).                                      *
 *                                                                             *
 * Written by Suiliang Qiu(Student Id=z3267009) for COMP9021                   *
 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * */

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
bool check_prime(int);
int main(void)
{
    for (int m = 100; m < 1000; ++m)
        for (int n = 10; n < 100; ++n) {
            int result = m * n;
            if (m * n >= 100000)
                break;
            int m1 = m / 100;
            int m2 = m / 10 % 10;
            int m3 = m % 10;
            int n1 = n / 10;
            int n2 = n % 10;
            //patial product
            int pp1 = m * n2;
            int pp2 = m * n1;
            if (pp1>7777&&pp2>7777&&result>77777)
                continue;
            //patial product digit
            int p11 = pp1 / 1000;
            int p12 = pp1 / 100 % 10;
            int p13 = pp1 / 10 % 10;
            int p14 = pp1 % 10;
            int p21 = pp2 / 1000;
            int p22 = pp2 / 100 % 10;
            int p23 = pp2 / 10 % 10;
            int p24 = pp2 % 10;
            //result digit
            int r1=result / 10000;
            int r2=result / 1000 % 10;
            int r3=result / 100 % 10;
            int r4=result / 10 % 10;
            int r5=result % 10;
            if (check_prime(m1)&&check_prime(m2)&&check_prime(m3)&&check_prime(n1)&&check_prime(n2)&&check_prime(p11)&&check_prime(p12)&&check_prime(p13)&&check_prime(p14)&&check_prime(p21)&&check_prime(p22)&&check_prime(p23)&&check_prime(p24)&&check_prime(r1)&&check_prime(r2)&&check_prime(r3)&&check_prime(r4)&&check_prime(r5))
                printf("%d x %d = %d, with %d, %d as partial products, is a solution.\n",
                   m, n, result, pp1, pp2);
        }
    
    return EXIT_SUCCESS;
}
bool check_prime(int i)
{
    if (i == 2 || i == 3 || i == 5 || i == 7)
        return true;
    return false;
}