ZOJ 3868(容斥原理+快速幂)

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 denotes E · (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
Submit    Status

题意：给出一个序列，求这个序列子集gcd的k次方和的期望

容斥原理。。。

#include<iostream>#include<algorithm>#include<cstdio>#include<vector>#include<cstring>#include<map>using namespace std;typedef long long ll;const int maxn = 1e6 + 10;const ll mod = 998244353;#define rep(i,a,b) for(int i=(a);i<(b);i++)#define pb push_backint cnt[maxn],a[maxn];ll d[maxn];ll quick_exp(ll a,int n){    ll res = 1;    while(n) {        if(n&1)res = res*a % mod;        a = a*a % mod;        n >>= 1;    }    return res;}int main(){        int T;        scanf("%d",&T);        while(T--) {            int n,k;            scanf("%d%d",&n,&k);            int Mgcd = 1;            for(int i = 0; i < n; i++) {                int x;                scanf("%d",&x);                Mgcd = max(Mgcd,x);                a[i] = x;            }            memset(cnt,0,sizeof(cnt[0])*Mgcd+20);            for(int i = 0; i < n; i++)cnt[a[i]]++;            ll ans = 0;            /*d[i] 表示子集的gcd为i的方案数*/            for(int i = Mgcd; i > 0; i--) {                int tot = 0;                for(int j = i; j <= Mgcd; j+= i)                {                    tot += cnt[j];                    d[i] = (d[i]-d[j])%mod;/*容斥*/                }                /*任选一个i的倍数的非空子集有2^tot-1种可能*/                d[i] = (d[i]+quick_exp(2,tot)-1)%mod;                d[i] = (d[i]+mod)%mod;                ans += d[i]*quick_exp(i,k);                ans %= mod;            }            cout<<ans<<endl;        }        return 0;}