Project Euler Problem 33

问题：

The fraction ⁴⁹/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that ⁴⁹/98 = ⁴/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, ³⁰/50 = ³/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

答案：100

#include
using namespace std;

int main(){

    int numd1, numd2, dend1, dend2;

    for(int num = 10; num < 100; ++num){
        for(int den = 10; den < 100; ++den){
            if(num % 10 == 0 && den % 10 == 0)continue;
            if(num > den)continue;
            numd1 = num /10;
            numd2 = num % 10;
            dend1 = den / 10;
            dend2 = den % 10;
            if(numd1 + numd2 == dend1 + dend2)continue;

            if((numd1 == dend1 && (numd2 * den == num * dend2))
               || (numd1 == dend2 && (numd2 * den == num * dend1))
               || (numd2 == dend1 && (numd1 * den == num * dend2))
               || (numd2  == dend2 && (numd1 * den == num * dend1))
               )cout << num << "," << den << endl;
        }
    }
    return 0;
}


输出：

16,64 19,95 26,65 49,98

1/4×1/5×2/5×4/8=1/100