141 lines
4.8 KiB
C++
141 lines
4.8 KiB
C++
// Source : https://leetcode.com/problems/maximize-number-of-nice-divisors/
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// Author : Hao Chen
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// Date : 2021-03-28
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/*****************************************************************************************************
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*
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* You are given a positive integer primeFactors. You are asked to construct a positive integer n that
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* satisfies the following conditions:
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*
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* The number of prime factors of n (not necessarily distinct) is at most primeFactors.
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* The number of nice divisors of n is maximized. Note that a divisor of n is nice if it is
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* divisible by every prime factor of n. For example, if n = 12, then its prime factors are [2,2,3],
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* then 6 and 12 are nice divisors, while 3 and 4 are not.
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*
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* Return the number of nice divisors of n. Since that number can be too large, return it modulo 10^9
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* + 7.
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*
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* Note that a prime number is a natural number greater than 1 that is not a product of two smaller
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* natural numbers. The prime factors of a number n is a list of prime numbers such that their product
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* equals n.
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*
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* Example 1:
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*
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* Input: primeFactors = 5
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* Output: 6
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* Explanation: 200 is a valid value of n.
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* It has 5 prime factors: [2,2,2,5,5], and it has 6 nice divisors: [10,20,40,50,100,200].
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* There is not other value of n that has at most 5 prime factors and more nice divisors.
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*
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* Example 2:
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*
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* Input: primeFactors = 8
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* Output: 18
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*
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* Constraints:
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*
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* 1 <= primeFactors <= 10^9
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******************************************************************************************************/
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/*
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considering `primeFactors = 5`
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So, we can have the following options:
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1) [2,3,5,7,11] - all of factors are different, then we only can have 1 nice divisor
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2) [2,2,3,5,7] - we can have 2*1*1*1 = 2 nice divisors: 2*3*5*7 and 2*2*3*5*7
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3) [2,2,3,3,5] - we can have 2*2*1 = 4 nice divisors: 2*3*5， 2*2*3*5， 2*3*3*5， 2*2*3*3*5
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4) [2,2,3,3,3] - we can have 2*3 = 6 nice divisors
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5）[2,2,2,2,3] - we can have 4*1 =4 nice divisors: 2*3, 2*2*3, 2*2*2*3, 2*2*2*2*3
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6) [2,2,2,2,2] - we can have 5 nice divisors: 2, 2*2, 2*2*2, 2*2*2*2, 2*2*2*2*2
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So, we can see we must have some duplicated factors.
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And what is the best number of duplication ?
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primeFactors = 1, then 1 - example: [2]
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primeFactors = 2, then 2 - example: [2,2]
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primeFactors = 3, then 3 - example: [5,5,5]
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primeFactors = 4, then 4 = 2*2 - example: [2,2,5,5])
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primeFactors = 5, then 5 = 2*3 - example: [3,3,3,5,5])
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primeFactors = 5, then 6 = 3*3 - example: [3,3,3,5,5,5])
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primeFactors = 7, then 3*4 = 12 - example: [3,3,3,5,5,5,5])
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primeFactors = 8, then 3*3*2 = 18 whcih > (2*2*2*2, 2*4*2, 3*5)
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primeFactors = 9, then 3*3*3 = 27
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primeFactors = 10, then 3*3*4 = 36
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So, we can see the '3' & '4' are specifial,
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- most of case, we can be greedy for `3`
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- but if the final rest is 4, then we need take 4.
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*/
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const int mod = 1000000007;
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class Solution {
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public:
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int maxNiceDivisors(int primeFactors) {
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return maxNiceDivisors_03(primeFactors);
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return maxNiceDivisors_02(primeFactors); //TLE
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return maxNiceDivisors_01(primeFactors); //TLE
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}
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int maxNiceDivisors_01(int primeFactors) {
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int result = 1;
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while ( primeFactors > 4 ) {
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primeFactors -= 3;
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result = (result * 3l) % mod;
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}
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result = (result * (long)primeFactors) % mod;
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return result;
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}
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int maxNiceDivisors_02(int primeFactors) {
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if (primeFactors <= 4 ) return primeFactors;
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int result = 1;
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for (int i = 4; i > 0; i-- ){
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if ((primeFactors - i) % 3 == 0){
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result = i;
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primeFactors -= i;
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// now, `primeFactors` is 3 times - 3X
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// we need convert 3X to 3^X
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for (int x = primeFactors/3; x > 0; x-- ) {
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result = (result * 3l) % mod;
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}
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break;
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}
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}
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return result;
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}
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int pow3(int x) {
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long result = 1;
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long factor = 3;
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while(x > 0) {
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if (x & 1) {
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result = (result * factor) % mod;
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}
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factor *= factor;
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factor %= mod;
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x /= 2;
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}
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return result % mod;
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}
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int maxNiceDivisors_03(int primeFactors) {
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if (primeFactors <= 4 ) return primeFactors;
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int result = 1;
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for (int i = 4; i > 0; i-- ){
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if ((primeFactors - i) % 3 == 0){
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primeFactors -= i;
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// now, `primeFactors` is 3 times - 3X
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// we need convert 3X to 3^X
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int x = primeFactors / 3;
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result = (long(i) * pow3(x)) % mod;
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break;
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}
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}
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return result;
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}
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};
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