GCD Expectation
Time Limit: 4 Seconds
Memory Limit: 262144 KB
Edward has a set of n integers {a1, a2,...,an}. He randomly picks a nonempty subset {x1,
x2,…,xm} (each nonempty subset has equal probability to be picked), and would like to know the expectation of [gcd(x1,
x2,…,xm)]k.
Note that gcd(x1, x2,…,xm) is the greatest common divisor of {x1,
x2,…,xm}.
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,…,an (1 ≤
ai ≤ 106).
The sum of values max{ai} 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
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,…,xm)]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 long
using 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;
}