cf 895C Square Subsets

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#bitmask #动归

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本文介绍了一道算法题目，任务是计算给定数组中元素的所有非空子集，找出那些子集的乘积能构成完全平方数的方法数量。通过使用动态规划和素数分解的思想，文章详细阐述了解决这一问题的具体步骤。

一原题

Petya was late for the lesson too. The teacher gave him an additional task. For some array a Petya should find the number of different ways to select non-empty subset of elements from it in such a way that their product is equal to a square of some integer.

Two ways are considered different if sets of indexes of elements chosen by these ways are different.

Since the answer can be very large, you should find the answer modulo 109 + 7.

Input

First line contains one integer n (1 ≤ n ≤ 105) — the number of elements in the array.

Second line contains n integers ai (1 ≤ ai ≤ 70) — the elements of the array.

Output

Print one integer — the number of different ways to choose some elements so that their product is a square of a certain integer modulo 109 + 7.

   Examples
  
     input
    
4
1 1 1 1

     output
    
15

     input
    
4
2 2 2 2

     output
    
7

     input
    
5
1 2 4 5 8

     output
    
7

Note

In first sample product of elements chosen by any way is 1 and 1 = 12. So the answer is 24 - 1 = 15.

In second sample there are six different ways to choose elements so that their product is 4, and only one way so that their product is 16. So the answer is 6 + 1 = 7.

二分析

题意：给定n个1到70之间的数，问有多少种组合方式，使得选出的数的乘积是完全平方数。

在[1, 70]区间上共有19个素数，很容易想到把整数i表示为一个19维的0-1向量，每一维表示有偶/奇数个对应的素因子，把这样的向量记作mask[i]。两个数相乘则对应为两个向量在每个维度上做异或，完全平方数对应全0向量。

理解了这一点之后，这道题就可以用动态规划解决了，dp[i][j](i<=70, j<2^20)表示考虑输入中小于等于i的数，这些数相乘得到j表示的向量的方案数。递推方程为：

dp[i+1][j] += dp[i][j] * f1[i+1], dp[i+1][j^mask[i+1]] += dp[i][j] * f2[i+1]

其中f1[i]表示从输入中选取偶数个i的方案数，f2[i]表示从输入中选取奇数个i的方案数。

这道动归的状态表达方法之前没见识过，RBQ RBQ

三代码

AC代码：

#include<iostream>
#include<cstring>
using namespace std;

const int inf = 72;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;

int n, times[inf];
bool prime[inf];
int mask[inf];
long long pow[maxn];
long long f1[inf], f2[inf];
long long dp[2][1 << 20]; // total 19 prime numbers in [1, 70]

void setMask() {
	for(int i = 1; i < inf; i++) {
		int cnt = 0;
		for(int j = 1; j <= i; j++) {
			if(!prime[j]) continue;
			int x = i;
			while(x % j == 0) {
				x /= j;
				mask[i] ^= (1 << cnt);
			}
			cnt++;
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	int val;
	for(int i = 0; i < n; i++) {
		cin >> val;
		times[val]++;
	}
	
	// prepare f1 and f2
	pow[0] = 1;
	for(int i = 1; i < maxn; i++) {
		pow[i] = (pow[i - 1] + pow[i -1]) % mod;
	}
	for(int i = 0; i < inf; i++) {
		if(times[i] == 0)
			f1[i] = 1, f2[i] = 0;
		else
			f1[i] = f2[i] = pow[times[i] - 1];
	}
	
	// prepare mask
	for(int i = 2; i < inf; i++) {
		prime[i] = true;
		for(int j = 2; j * j <= i; j++) {
			if(i % j == 0) {
				prime[i] = false;
				break;
			}
		}
	}
	setMask();

	// begin dp
	dp[0][0] = 1;
	for(int i = 1; i <= 70; i++) {
		int cur = i % 2;
		memset(dp[cur], 0, sizeof(dp[cur]));
		for(int j = 0; j < (1 << 20); j++) {
			dp[cur][j] += (dp[1 - cur][j] * f1[i]) % mod;
			dp[cur][j] %= mod;
			dp[cur][j ^ mask[i]] += (dp[1 - cur][j] * f2[i]) % mod;
			dp[cur][j] %= mod;
		}
	}
	
	cout << (dp[0][0] - 1) % mod << endl; // due to 70 is even, it's dp[0][0]

	return 0;
}