ZOJ-3868-GCD Expectation（容斥）

GCD Expectation

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Time Limit: 4 Seconds      Memory Limit: 262144 KB

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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 denotesE · (2ⁿ - 1) modulo 998244353.

Sample Input

15 11 2 3 4 5

Sample Output

42

题目连接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5480

题意 ，给出序列a[1],a[2]...a[i],求子序列 [gcd(x₁, x₂,…,x_m)]^k.   的期望乘  (2ⁿ - 1)  ，   显然子序列个数有 (2ⁿ - 1) 个，那么其实就是所有子序列gcd的k次方求和啦

数据范围不超1e6，那么可以直接枚举gcd就好啦，dp[i]表示gcd为i的组数，从高到低枚举gcd，计算出i的倍数出现了m次，那么就有2^m-1组为i倍数的子序列，

显然，dp[i]=2^m-1-dp[i*2]-dp[i*3]-....;

//#include<bits/stdc++.h>#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <cstring>#define LL long longusing namespace std;const int MAXN = 1e6+7;const int MOD = 998244353;long long a[MAXN],dp[MAXN];long long quick_pow(long long a,long long n){    long long ans=1;    while(n)    {        if(n&1)ans*=a,ans%=MOD;        a=a*a%MOD;        n>>=1;    }    return ans;}int main(){    int T,n,k,temp;    cin>>T;    int cas=1;    while(T--)    {        memset(a,0,sizeof(a));        memset(dp,0,sizeof(dp));        scanf("%d%d",&n,&k);        int maxx=0,x;        for(int i=0; i<n; ++i)        {            scanf("%d",&x);            a[x]++;            maxx=max(maxx,x);        }        long long ans=0;        for(int i=maxx; i>=1; --i)        {            int cnt=0,temp=0;            for(int j=i; j<=maxx; j+=i)            {                cnt+=a[j];                temp=(temp-dp[j]+MOD)%MOD;            }            dp[i]=((quick_pow(2,cnt)-1)%MOD+temp)%MOD;            ans=(ans+dp[i]*quick_pow(i,k)%MOD)%MOD;        }        cout<<ans<<endl;    }    return 0;}