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.
OutputFor each case, if the expectation is E, output a single integer denotes E · (2n - 1) modulo 998244353.
Sample Input1 5 1 1 2 3 4 5Sample Output
42
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=1e6+10; const int inf=0x3f3f3f3f; const int mod=998244353; ll num_mul[maxn],sum[maxn]; int val[maxn]; /* 数量也必须同时取余,因为后面有可能导致这个值变得很“大 ”,导致没有办法计算 利用容斥原理,感觉更像是筛选,i 去掉i的倍数 */ void init(){ memset(num_mul,0,sizeof(num_mul)); memset(sum,0,sizeof(sum)); } ll q_pow(ll base,int n){ ll ans=1; while(n){ if(n&1){ ans=(ans*base)%mod; } base=(base*base)%mod; n>>=1; } return ans; } int main(){ int T; scanf("%d",&T); while(T--){ init(); int n,k,mv=0; scanf("%d %d",&n,&k); for(int i=1;i<=n;i++){ scanf("%d",&val[i]); sum[val[i]]++; mv=max(mv,val[i]); } ll ans=0; for(int i=mv;i>=1;i--){ int cnt=0; for(int j=i;j<=mv;j+=i){ cnt+=sum[j]; num_mul[i]=(num_mul[i]-num_mul[j]+mod)%mod;; } num_mul[i]=(num_mul[i]+q_pow((ll)2,cnt)-1)%mod; ans=(ans+num_mul[i]*q_pow((ll)i,k))%mod; } printf("%lld\n",ans); } return 0; }