ZOJ 3868 GCD Expectation (容斥＋莫比乌斯反演)

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 integerT 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 containsn integersa₁, 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

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Author: LIN, Xi
Source: The 15th Zhejiang University Programming Contest

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

题目大意：给一个集合，{xi}为它的一个非空子集，设E为[gcd(x₁,x₂,…,x_m)]^k  的期望，求E*(2^n - 1) mod 998244353

题目分析：首先一个有n个元素的集合的非空子集个数为2^n - 1，所以E的分母就是2^n - 1了，因此我们要求的只是E的分子,

设F(x)为gcd(xi) = x的个数，那么ans = (1^k) * F(1) + (2^k) * F(2) + ... + (ma^k) * F(ma)

下面的问题就是如何快速的计算F(x)了，对于一个集合，先计算出x的倍数的个数，nlogn即可，然后就是基础的容斥，假设现在要求gcd为1的，那就减去gcd为2的，gcd为3的，注意到6同时是2和3的倍数，也就是6的倍数被减了两次，所以要加上gcd为6的，前面的系数刚好是数字对应的莫比乌斯函数，看到这题很多用dp来容斥的，其实本质和莫比乌斯函数一样，但是莫比乌斯函数写起来真的很简单，2333333

#include <cstdio>#include <cstring>#include <algorithm>#define ll long longusing namespace std;int const MOD = 998244353;int const MAX = 1e6 + 5;ll two[MAX];int p[MAX], mob[MAX], num[MAX], cnt[MAX];bool noprime[MAX];int n, k, ma, pnum;void Mobius(){    pnum = 0;    mob[1] = 1;    for(int i = 2; i < MAX; i++)    {        if(!noprime[i])        {            p[pnum ++] = i;            mob[i] = -1;        }        for(int j = 0; j < pnum && i * p[j] < MAX; j++)        {            noprime[i * p[j]] = true;            if(i % p[j] == 0)            {                mob[i * p[j]] = 0;                break;            }            mob[i * p[j]] = -mob[i];        }    }}ll qpow(ll x, ll n){    ll res = 1;    while(n != 0)    {        if(n & 1)            res = (res * x) % MOD;        x = (x * x) % MOD;        n >>= 1;    }    return res;}void pre(){    Mobius();       two[0] = 1;    for(int i = 1; i < MAX; i++)        two[i] = two[i - 1] * 2ll % MOD;}int main(){    pre();    int T;    scanf("%d", &T);    while(T --)    {        memset(num, 0, sizeof(num));        memset(cnt, 0, sizeof(cnt));        ma = 0;        int tmp;        scanf("%d %d", &n, &k);        for(int i = 0; i < n; i++)        {            scanf("%d", &tmp);            cnt[tmp] ++;            ma = max(ma, tmp);        }        for(int i = 1; i <= ma; i++)            for(int j = i; j <= ma; j += i)                num[i] += cnt[j];       //求i的倍数的个数        ll ans = 0;        for(int i = 1; i <= ma; i++)    //枚举gcd        {            ll sum = 0;            for(int j = i; j <= ma; j += i)  //容斥                sum = (MOD + sum % MOD + mob[j / i] * (two[num[j]] - 1) % MOD) % MOD;            ans = (MOD + ans % MOD + (sum * qpow(i, k)) % MOD) % MOD;        }        printf("%lld\n", ans);    }}