GCD Expectation ZOJ - 3868 (Inclusion and Exclusion Principle + Fast Power)

GCD Expectation

ZOJ - 3868

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

Title:

Given n numbers {a1,a2,a3...an}, let us sum the k-th power of the greatest common divisor of any subset of this set

Ideas:

We can convert the problem into finding the number of non-empty sets with gcd equal to i, so that we can calculate directly.

We store the largest of the n numbers as Max when inputting, because the greatest common factor of these n numbers must not exceed Max, and record the number of each number

gcd i enumerates from Max to 1

For each greatest common factor i, we ask for the number of multiples of i. The inner loop starts from i and increases exponentially to Max +i each time, because we have recorded the number of each number. , nothing is zero, so the number of multiples of i can be obtained by directly adding in the loop process

Suppose there are x numbers of n numbers that are multiples of i.

Then the number of gcd equal to i is the total (2^x-1) (all sets with i as a multiple) minus the number of gcd equal to j, j is a multiple of i (that is, the greatest common factor is i number of multiples), which is equivalent to the simplest case of the inclusion-exclusion theorem.

After the last gcd is the number of i dp[i], directly find dp[i]*q_pow(i,k) and add it to the answer to take the modulo

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int maxn = 1e6+100;
int sum[maxn];//The number of each number given
int a[maxn];//The given array
int dp[maxn];//dp[i] is the number of numbers whose i is the greatest common factor
ll q_pow(ll a,ll b){
    ll ans = 1;
    while(b){
        if(b & 1)
            years = years * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return ans;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        memset(sum,0,sizeof(sum));
        memset(dp,0,sizeof(dp));
        int n,k,Max = 0;
        scanf("%d%d",&n,&k);
        for(int i = 0; i < n; i++){
            scanf("%d",&a[i]);
            Max = max(a[i],Max);
            sum[a[i]]++;
        }
        ll ans = 0;
        for(int i = Max; i >= 1; i--){
            int cnt = 0;
            for(int j = i; j <= Max; j += i){
                cnt += sum[j];//How many multiples of i are recorded
                dp[i] = (dp[i] - dp[j] + mod) % mod;
                //Calculate how many numbers have i as the greatest common factor, subtract the total set of multiples of i
                //The number of the greatest common factor which is a multiple of i
            }
            dp[i] = ((dp[i] + q_pow(2,cnt) - 1) % mod + mod) % mod;
            ans = (ans + dp[i] * q_pow(i,k)) % mod;//Calculate the k power of gcd, and there are dp[i]
        }

        printf("%lld\n",ans);
    }
    return 0;
}