What is the minimum number of factors in the inverse prime number template?

Antiprime Number:

If a natural number is more than the number of all natural numbers smaller than it, then we call this number a antiprime number. For example, values 1, 2, 4, 6, 12, and 24 are inverse prime numbers.

 

The maximum antiprime Number of N is as follows:

# Include <math. h >#include <iostream> using namespace STD; # define ll _ int64const LL Prime [16] = {1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}; ll maxsum, bestnum, N; void DFS (LL num, ll K, ll sum, ll limit) // num: the number currently enumerated. K: the prime factor of the K enumerated. Sum: the approximate number of the numbers. Limit: the maximum number of quality factors; {int I; ll TMP; If (sum> maxsum) {maxsum = sum; bestnum = num; // if there are more approx., update the optimal solution to the current number ;} if (sum = maxsum & bestnum> num) {// If the approximate number is the same, update the optimal solution to a smaller number; bestnum = num;} If (k> 15) return; TMP = num; for (I = 1; I <= limit; ++ I) {// start to enumerate the number of each prime factor if (TMP * prime [k]> N) {break;} TMP * = prime [k]; // accumulate to the current number DFS (TMP, k + 1, sum * (I + 1), I); // continue the next search} int main () {ll I, j; int N; CIN> N; DFS (1, 1, 1, 50); printf ("% LLD \ n", bestnum ); return 0 ;}

The following table can be used to calculate 1-1000 or more inverse prime numbers (just change the array size)

# Include <iostream> # include <cstdio> # include <algorithm> # include <cmath> # include <cstring> using namespace STD; typedef _ int64 LLD; lld p [1010]; // P [I] indicates the smallest integer of the number of factors I. What is LLD prime [30] = {2, 3, 5, 7, 11, 13, 17, 19, 29, 31,37, 41,43, 47,53}; int maxn; // the maximum possible value void getartprime (LLD cur, int CNT, int limit, int K) {// cur: the number currently enumerated; // CNT: the number of factors involved; // limit: the upper limit of the number of factors; 2 ^ T1 * 3 ^ T2 * 5 ^ T3 ...... T1> = T2> = T3 ...... Generally, 50 is enough. // The Prime Number of K. // If (cur> (LLD) 1 <60) | CNT> 150) return; if (cur> maxn) return; // if the current number is greater than the maximum number we require, maxn will be able to search for it. If (P [CNT]! = 0 & P [CNT]> cur) // The number of current factors has been recorded and the number of records at that time is greater than the number currently enumerated, then, replace the enumerated number P [CNT] = cur under the number of this factor; If (P [CNT] = 0) // the number of this factor has not appeared, P [CNT] = cur; LLD temp = cur; For (INT I = 1; I <= limit; I ++) // enumeration count {temp = temp * prime [k]; If (temp> maxn) return; getartprime (temp, CNT * (I + 1), I, k + 1) ;}} int main () {getartprime (, 50, 0); Return 0 ;}