ZOJ 2562 more divisors (high synthesis number)

ZOJ 2562 more divisors (high synthesis number)

ACM

Title Address: ZOJ 2562 more divisors

Test Instructions :
Find the largest high synthetic number less than n, a high synthetic number refers to a class of integers, no matter what the number of natural number is smaller than the number of factors.

Analysis :
On the internet called it anti-Prime, in fact I checked, the number of the number should be written backwards write is the prime number of prime. This is supposed to be called a high synth, see wikipedia:highly composite number.

The high composition number has the following characteristics:
where p 1 < p 2 <?< p k Is prime, and the exponents c i is positive integers.
Any factor of n must has the same or lesser multiplicity in each prime:
P1^d1x P2^d 2 x?x Pk^d k , 0≤ d i ≤ c i , 0<I≤k
So use backtracking enumeration.

Code :

/** Author:illuz <iilluzen[at]gmail.com>* blog:http://blog.csdn.net/hcbbt* file:2562.cpp* Create date:2014-08-06 20:45:53* descripton:highly Composite number*/#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace std; #define REPF (I,A,B) for (int i= (a); i<= (b); i++ typedef long LONG ll;const int M = 1000;ll n;ll bestnum;ll bestsum;ll hcm[m][2];ll prim[] = {2, 3, 5, 7, 11, 13, 17, 19,  23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67}; Current is num, with the prim[k], sum of divisors, the limit of prim[k] you can usevoid getnum (ll num, int k, ll sum, in T limit) {if (sum > bestsum) {bestsum = Sum;bestnum = num;} else if (sum = = bestsum && num < bestnum) {BESTN Um = num;}  ll p = prim[k];for (int i = 1; I <= limit; i++, p *= prim[k]) {//Use I prim[k]sif (num * p > N) break;getnum (num *= Prim[k], k + 1, Sum * (i + 1), i);}} Clac log2 (n) int log2 (ll n) {int ret = 0;ll p = 1;while (P < n) {p <<= 1;ret++;} return ret;} Return the number of highly Composite number in [1, n]//and save the HCM in Hcm[][2]int gethcm () {int ret = 0;n = 5000  00;//[1, n]while (n > 0) {bestnum = 1;bestsum = 1;getnum (1, 0, 1, log2 (n)); cout << bestnum << ' << Bestsum << endl;hcm[ret][0] = bestnum;hcm[ret][1] = Bestsum;n = bestnum-1;ret++;} return ret;} int main () {while (CIN >> n) {bestnum = 1;bestsum = 1;getnum (1, 0, 1, +); cout << bestnum << Endl;}}

ZOJ 2562 more divisors (high synthesis number)