GCD Expectation ZOJ - 3868 (brute force solution)

Edward has a set of n integers {a₁, a₂,...,a_n}. He randomly picks a nonempty subset {x₁, x₂,…,x_m} (each nonempty subset has equal probability to be picked), and would like to know the expectation of [gcd(x₁, x₂,…,x_m)]^k.

Note that gcd(x₁, x₂,…,x_m) is the greatest common divisor of {x₁, x₂,…,x_m}.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers n, k (1 ≤ n, k ≤ 10⁶). The second line contains n integers a₁, a₂,…,a_n (1 ≤ a_i ≤ 10⁶).

The sum of values max{a_i} for all the test cases does not exceed 2000000.

Output

For each case, if the expectation is E, output a single integer denotes E · (2ⁿ - 1) modulo 998244353.

Sample Input

1
5 1
1 2 3 4 5

Sample Output

42 

The meaning of the question: Given n and k, find the expectation of the k-th power of any non-empty subset gcd of n numbers, and finally multiply the expectation by 2^n-1
Thought: because each subset is equal probability, take each set The probability of is 1/(2^n-1), so in the end, what is actually calculated is the sum of the k-th power of gcd of each non-empty set.
We can convert the problem to find how many non-empty sets with gcd equal to i are there . gcd enumerates from 1 to the maximum maxt to find the answer.
For each i, let there be x number of n numbers that are multiples of i.
Then, the number of gcd equal to i is equal to the total of 2^x-1 minus gcd is equal to the number of j, j is a multiple of i
namely : dp[i]=2^x-1-dp[i*2]-dp[i*3] - .......,dp[i ]b indicates the number of sets whose gcd is i

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define mod 998244353
using namespace std;
typedef long long ll;
const int N = 2000200;
int sum[N],arr[N];
int dp[N];
ll quick(ll a,int b){
	ll ans=1;
	while(b){
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
	    memset(sum,0,sizeof(sum));
		memset(dp,0,sizeof(dp));
		int n,k,maxt=0;
		scanf("%d%d",&n,&k);
		for(int i=0;i<n;i++){
			scanf("%d",&arr[i]);
			maxt=max(maxt,arr[i]);
			sum[arr[i]]++; //The number of occurrences of each number
		}
		ll ans=0;
		for(int i=maxt;i>=1;i--){
			int tmp=0;
			for(int j=i;j<=maxt;j+=i){
				tmp+=sum[j];
				dp[i]=(dp[i]-dp[j]+mod)%mod;
			}
			dp[i]=((dp[i]+quick(2,tmp)-1)%mod+mod)%mod;
			ans=(ans+dp[i]*quick(i,k))%mod;
		}
		printf("%lld\n",ans);
	}
	return 0;
}