The Number of Good Subsets

The Number of Good Subsets

You are given an integer array nums . We call a subset of nums good if its product can be represented as a product of one or more distinct prime numbers.

• For example, if nums = [1, 2, 3, 4] :
  
  • [2,3] ,  [1, 2, 3] , and [1, 3] are good subsets with products 6 = 2*3 , 6 = 2*3 , and 3 = 3 respectively.
  • [1, 4] and [4] are not good subsets with products 4 = 2*2 and 4 = 2*2 respectively.

Return the number of different good subsets in nums modulo ${10}^9 + 7$.

A subset of nums is any array that can be obtained by deleting some (possibly none or all) elements from nums . Two subsets are different if and only if the chosen indices to delete are different.

Example 1:

Input: nums = [1,2,3,4]
Output: 6
Explanation: The good subsets are:
- [1,2]: product is 2, which is the product of distinct prime 2.
- [1,2,3]: product is 6, which is the product of distinct primes 2 and 3.
- [1,3]: product is 3, which is the product of distinct prime 3.
- [2]: product is 2, which is the product of distinct prime 2.
- [2,3]: product is 6, which is the product of distinct primes 2 and 3.
- [3]: product is 3, which is the product of distinct prime 3.

Example 2:

Input: nums = [4,2,3,15]
Output: 5
Explanation: The good subsets are:
- [2]: product is 2, which is the product of distinct prime 2.
- [2,3]: product is 6, which is the product of distinct primes 2 and 3.
- [2,15]: product is 30, which is the product of distinct primes 2, 3, and 5.
- [3]: product is 3, which is the product of distinct prime 3.
- [15]: product is 15, which is the product of distinct primes 3 and 5.

Constraints:

$1 \leq \text{nums.length} \leq 10^5$
$1 \leq \text{nums}[i] \leq 30$

 

解题思路

　　Count the Number of Square-Free Subsets的扩展版，$n$扩大到了$10^5$，当然题意也有些变化（港真lc直接把原题搬到周赛这样做真的好吗）。如果还是像这题一样的做法，由于时间复杂度为$O(n \cdot 2^{10})$，那么一定会超时。

　　注意，由于每个数最大不超过$30$，因此我们可以把值域定义为状态，而不是把下标定义为状态。

　　定义状态$f(i,j)$表示从数值$2 \sim i$中选择，乘积结果的状态为$j$（即包含了哪些质数）的所有方案的个数。其中根据数值$i$选或不选来划分状态。状态转移方程为 $$f(i,j) = f(i-1,j) + f(i-1,j \wedge st_{i}) \times \text{cnt}_i \ \ \ \ \ (\text{保证} j \ \& \ st_{i} = st_i)$$

　　其中$\text{cnt}_i$表示数值$i$出现的次数（显然一个合法的结果的乘积只能乘一次$i$，一次有多少个$i$就有多少种方案）。如果$i$本身就不合法（即某个质因子次数超过$1$），那么很明显该数是不能选择的。为了方便这里把不合法的数的$\text{cnt}_i$定义为$0$，那么同样适用上面的状态转移方程。

　　为什么要从$2$开始呢？$1$的情况呢？该题的要求是“所有元素的乘积可以表示为一个或多个互不相同的质数的乘积”，而$1$不是质数，因此乘积的结果不能是$1$，但可以用$1$去乘其他的数。我们把$1$归为状态$0$，显然$f(1,0) = 2^{\text{cnt}_1}$，其他的$i \ (2 \leq i \leq 30)$可以从$f(1,0)$转移过来。

　　最后的答案就是$\sum\limits_{i=1}^{2^{10}-1}{f(30,i)}$，$i$不从$0$开始是为了不统计乘积结果为$1$的情况。

　　AC代码如下，时间复杂度为$O(30 \cdot 2^{10})$：

class Solution {
public:
    int mod = 1e9 + 7;
    
    int numberOfGoodSubsets(vector<int>& nums) {
        int primes[10] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};
        vector<int> cnt(31), st(31);
        for (auto &x : nums) {
            bool flag = true;
            int t = x;
            for (int i = 0; i < 10; i++) {
                int p = primes[i];
                if (t % p == 0) {
                    int s = 0;
                    st[x] |= 1 << i;
                    while (t % p == 0) {
                        s++;
                        t /= p;
                    }
                    if (s > 1) {
                        flag = false;
                        break;
                    }
                }
            }
            if (flag) cnt[x]++;
        }
        vector<vector<int>> f(31, vector<int>(1 << 10));
        f[1][0] = 1;
        while (cnt[1]--) {
            f[1][0] = f[1][0] * 2 % mod;
        }
        for (int i = 2; i <= 30; i++) {
            for (int j = 0; j < 1 << 10; j++) {
                f[i][j] = f[i - 1][j];
                if ((j & st[i]) == st[i]) f[i][j] = (f[i][j] + 1ll* f[i - 1][j ^ st[i]] * cnt[i]) % mod;
            }
        }
        int ret = 0;
        for (int i = 1; i < 1 << 10; i++) {
            ret = (ret + f[30][i]) % mod;
        }
        return ret;
    }
};

 

参考资料

　　好子集的数目：https://leetcode.cn/problems/the-number-of-good-subsets/solution/hao-zi-ji-de-shu-mu-by-leetcode-solution-ky65/