int main() {
int n = 10000;
std::cout << "Trailing 0s of " << n << "! is "
<< trailingZeroes(n) << std::endl;
assert(2499 == trailingZeroes(n));
return 0;
}
TEST(FactorialTrailingZeroesTestCase, Normal) {
EXPECT_EQ(1, trailingZeroes(5));
EXPECT_EQ(2, trailingZeroes(10));
EXPECT_EQ(6, trailingZeroes(25));
EXPECT_EQ(452137076, trailingZeroes(1808548329));
std::cout << "int " << sizeof(int) << std::endl;
std::cout << "unsigned int " << sizeof(unsigned int) << std::endl;
std::cout << "long " << sizeof(long) << std::endl;
std::cout << "long long " << sizeof(long long) << std::endl;
}
int trailingZeroes(int n) {
return n>0 ? n/5+trailingZeroes(n/5) : 0;
}
//recursive solution, code from oj leetcode discussion, a little faster than iterative solution
int trailingZeroes2(int n) {
return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
}
int main(int argc, char *argv[])
{
printf("%d\n", trailingZeroes(atoi(argv[1])));
return(0);
}
int trailingZeroes(int n) {
return (n == 0)? 0: n / 5 + trailingZeroes(n/5);
}
// all trailing 0 is from factors 5 * 2
// and number of factor 2 is more than number of factor 5
// calculate number of 5 and exponentials of 5 from 1 to n
// use division rather than multyply can avoid overflow
int trailingZeroes(int n) {
return n < 5 ? 0 : n / 5 + trailingZeroes(n / 5);
}
