313. Super Ugly Number

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.

这个题目与丑数类似，只是增加了候选公约数的数目.将原来的候选数组改成矩阵。

思路1：下面的编程非常巧妙，去除了重复因子

class Solution {public:    int nthSuperUglyNumber(int n, vector<int>& primes) {        vector<int> dp(n,0);        vector<int> idx(primes.size(),0);//保存每个候选因子的当前乘积下标        vector<int> val(primes.size(),1);//保存当前因子的乘积        dp[0]=1;        int next=1;                for(int i=0;i<n;i++)        {            dp[i]=next;            next=numeric_limits<int>::max();            for(int j=0;j<primes.size();j++)            {                if(dp[i]==val[j]) val[j]=primes[j]*dp[idx[j]++];                next=min(next,val[j]);            }        }        return dp[n-1];            }};

思路2:使用优先级队列

class Solution {public:    int nthSuperUglyNumber(int n, vector<int>& primes) {        priority_queue<num,vector<num>,comp> q;        for(int i=0;i<primes.size();i++)        {            q.emplace(primes[i],1,primes[i]);        }        vector<int> res(n,0);        res[0]=1;        for(int i=1;i<n;i++)        {            res[i]=q.top().val;            while(q.top().val==res[i])            {                num temp=q.top();                q.pop();                q.emplace(res[temp.idx]*temp.p,temp.idx+1,temp.p);            }        }        return res[n-1];                    }    struct num{      int val;//当前因子的下一个候选值        int idx;//当前因子的下标        int p;//当前因子        num(int a,int b,int c):val(a),idx(b),p(c){                    }    };    struct comp{      bool operator()(const num& a,const num& b){          return a.val>b.val;      }      };};