Description

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.

Example:

Input:

n = 12, primes = [2,7,13,19]

Output:

32

Explanation:

[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 is a super ugly number for any given primes.
  • The given numbers in primes are in ascending order.
  • 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
  • The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

分析

题目的意思是：求一个数的第n个超级丑数。

  • 由于不知道质数的个数，可以用一个idx数组来保存当前的位置，然后我们从每个子链中取出一个数，找出其中最小值，然后更新idx数组对应位置，注意有可能最小值不止一个，要更新所有最小值的位置。

代码

class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        vector<int> dp(n,INT_MAX);
        vector<int> count(primes.size(),0);
        dp[0]=1;
        for(int i=1;i<n;i++){
            for(int j=0;j<primes.size();j++){
                dp[i]=min(dp[i],dp[count[j]]*primes[j]);
            }
            for(int j=0;j<primes.size();j++){
                if(dp[i]==dp[count[j]]*primes[j]){
                    count[j]++;
                }
            }
        }
        return dp[n-1];
    }
};

参考文献

​​[LeetCode] Super Ugly Number 超级丑陋数​​