Ultraviolet A Problem 10139 factovisors (factorial and Division)

// Factovisors (factorial and division) // PC/Ultraviolet IDs: 110704/10139, popularity: A, success rate: average level: 2 // verdict: accepted // submission date: 2011-06-11 // UV Run Time: 0.312 S // copyright (c) 2011, Qiu. Metaphysis # Yeah dot net // The idea is relatively simple. Calculate the number M and number 2-n gcd. If GCD is greater than 1, m/= GCD, if the last M is not // 1, it indicates that M cannot divide N !, Note that the input may exist in some special cases. You only need to handle it. It is strange that on the // programming challenges, the Rt is 0.072 seconds, but on the ultraviolet A, it is always TLE. Then, a statement for testing the prime number is added to the // command, it should be caused by special test data. Therefore, you need to check the conditions in the loop in time and exit the // loop to avoid TLE. # Include <iostream> # include <cmath> using namespace STD; // determine whether a number is a prime number. Bool is_prime (long unsigned number) {If (number = 2) return true; If (Number % 2 = 0) return false; For (long unsigned c = 3; c <= Ceil (SQRT (number); C + = 2) if (Number % C = 0) return false; return true;} // calculate the two-digit GCD. Long unsigned gcd (long unsigned A, long unsigned B) {if (a <B) {long unsigned TMP = A; A = B; B = TMP ;} while (B> 0) {long unsigned TMP = A; A = B; B = TMP % B;} return a;} int main (INT AC, char * AV []) {long unsigned n, m; // read N and M. While (CIN> N> m) {bool divided = false; long unsigned T = m, G; // in special cases, M = 1. If (M = 1) {divided = true; goto print;} // in special cases, M = 1. If (M = 0) goto print; // If M is a prime number and M is greater than N, M and 2-N are mutually Prime. If (is_prime (m) & M> N) goto print; // m <= N, of course, division can be performed. If (M <= N) {divided = true; goto print;} // you can find gcd (M, c) in descending order of N to 2 ). For (long unsigned c = N; C> = 2; c --) {G = gcd (M, c); // when there is a maximum public approx greater than 1, determine whether exit conditions are met. If (G> 1) {M/= g; // m <C, division is allowed. If (M <c) {divided = true; goto print;} // m> = C. If (is_prime (M) goto print ;}// print the result. Note the differences in output words. Print: If (divided) cout <t <"divides" <n <"! "<Endl; elsecout <t <" does not divide "<n <"! "<Endl;} return 0 ;}