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 sizek. 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. Credits:
Special thanks to for adding this problem and creating all test cases.这道题让我们求超级丑陋数，是之前那两道和的延伸，质数集合可以任意给定，这就增加了难度。但是本质上和没有什么区别，由于我们不知道质数的个数，我们可以用一个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，参见代码如下：
解法一：
class Solution {public: int nthSuperUglyNumber(int n, vector & primes) { vector res(1, 1), idx(primes.size(), 0); while (res.size() < n) { vector tmp; int mn = INT_MAX; for (int i = 0; i < primes.size(); ++i) { tmp.push_back(res[idx[i]] * primes[i]); } for (int i = 0; i < primes.size(); ++i) { mn = min(mn, tmp[i]); } for (int i = 0; i < primes.size(); ++i) { if (mn == tmp[i]) ++idx[i]; } res.push_back(mn); } return res.back(); }}; 上述代码可以稍稍改写一下，变得更简洁一些，原理完全相同，参见代码如下：
解法二：
class Solution {public: int nthSuperUglyNumber(int n, vector & primes) { vector dp(n, 1), idx(primes.size(), 0); for (int i = 1; i < n; ++i) { dp[i] = INT_MAX; for (int j = 0; j < primes.size(); ++j) { dp[i] = min(dp[i], dp[idx[j]] * primes[j]); } for (int j = 0; j < primes.size(); ++j) { if (dp[i] == dp[idx[j]] * primes[j]) { ++idx[j]; } } } return dp.back(); }}; 本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。