Write a program to find the nth super ugly number.

Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4.

Note:
​ (1) 1 is a super ugly number for any given primes.
​ (2) The given numbers in primes are in ascending order.
​ (3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
​ (4) The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

思路

在这道题，我们可以仿照Ugly Number II的解法来解决。

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class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        int p = int(pow(2.0,32.0) - 1.0);
        vector<double> uglynumber;
        uglynumber.push_back(1.0);
        vector<double> primespointer(primes.size());
        vector<double> tmpprimes(primes.size());
        while(n > 1){
            int maxint = p;
            for(int i = 0; i < primes.size(); i++){
                tmpprimes[i] = uglynumber[primespointer[i]] * double(primes[i]);
            }
            double min = p;
            double target = 0;
            for(int i = 0; i < primes.size(); i++){
                if(tmpprimes[i] < min){
                    min = tmpprimes[i];
                    target = i;
                }
            }
            primespointer[target]++;
            if(min != uglynumber.back()){
                uglynumber.push_back(min);
                n--;
            }
        }
        return uglynumber.back();
    }
};