ZOJ 3868 GCD Expectation

GCD Expectation

Time Limit: 4000ms

Memory Limit: 262144KB

This problem will be judged on ZJU. Original ID: 3868
64-bit integer IO format: %lld      Java class name: Main

 

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

Source

The 15th Zhejiang University Programming Contest

Author

LIN, Xi

 

解题：我们采取的是枚举gcd的策略

真子集的个数为$2^{cnt}-1$个数

但是我们发现，算我们枚举的gcd的时候，真正的gcd确实我们枚举的gcd的倍数

所以要去重，利用容斥原理

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod = 998244353;
const int maxn = 1000010;
LL dp[maxn];
int sum[maxn];
LL quickPow(LL base,int index){
    LL ret = 1;
    while(index){
        if(index&1) ret = ret*base%mod;
        index >>= 1;
        base = base*base%mod;
    }
    return ret;
}
int main(){
    int kase,n,k,mxv,tmp;
    scanf("%d",&kase);
    while(kase--){
        scanf("%d%d",&n,&k);
        memset(sum,0,sizeof sum);
        for(int i = mxv = 0; i < n; ++i){
            scanf("%d",&tmp);
            sum[tmp]++;
            mxv = max(mxv,tmp);
        }
        LL ret = 0;
        for(int i = mxv; i > 0; --i){
            int cnt = dp[i] = 0;
            for(int j = i; j <= mxv; j += i){
                dp[i] = (dp[i] - dp[j] + mod)%mod;
                cnt += sum[j];
            }
            dp[i] = (dp[i] + quickPow(2,cnt) - 1)%mod;
            ret = (ret + dp[i]*quickPow(i,k))%mod;
        }
        printf("%lld
",ret);
    }
    return 0;
}