313. Super Ugly Number - 思路详解

题目

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.

题目翻译

超级素数，素因子只包含规定素因子序列的元素。然后求第n个超级素数。

注意：
1，1是超级素数
2，给的素因子序列为升序

分析

该题可以任务是丑数的一个变种。只是素因子变为了一个素数序列。本质没有变。所以思路请出门右转，看
264. Ugly Number II-思路详解 http://blog.youkuaiyun.com/SJWL2012/article/details/53192810

代码

class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        int primes_size = primes.size();
        int *pUglyNumber  = new int[n];

        pUglyNumber[0] = 1;
        int nextUglyIndex = 1;

        vector<int*> pMultiN(primes_size,pUglyNumber);    //初始化

        while(nextUglyIndex < n){
            int min = FindMin(pMultiN,primes);

            pUglyNumber[nextUglyIndex] = min;

            for(int i = 0; i < primes_size; i ++){
                while(*(pMultiN[i])*primes[i] <= pUglyNumber[nextUglyIndex]){
                    ++pMultiN[i];
                }
            }

            ++nextUglyIndex;
        }

        int ugly = pUglyNumber[n-1];
        delete[] pUglyNumber;

        return ugly;
    }


    int FindMin(vector<int*> &pMultiN,vector<int> &primes){
        int primes_size = primes.size();
        int min = *(pMultiN[0])*primes[0];

        for(int i = 1; i < primes_size; ++i){
            if(*(pMultiN[i])* primes[i] < min){
                min = *(pMultiN[i])* primes[i];
            }
        }
        return min;
    }

};