zoj[3868]gcd期望

题意：求n个数组成的集合的所有非空子集的gcd的期望

大致思路：对于一个数x，设以x为约数的数的个数为cnt[x]，所组成的非空集合个数有2^cnt[x]-1个，这其中有一些集合的gcd是x的倍数的，怎么求得最终结果呢？下面来说明过程。

令f[x] = 2^cnt[x]-1，表示以x为gcd的集合个数。令maxn为所有数的最大值，一开始f[maxn]=2^cnt[maxn]-1是肯定正确的。若从大到小更新f数组，类似数学归纳法，f[x]需要减去f[2x]、f[3x]、...、f[px]，px<=maxn，而f[2x]、f[3x]、...、f[px]都是正确的，所以f[x]也是正确的。所以可以得到正确的f数组，有了f数组，答案自然出来了。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a);
#define sint(a) ReadInt(a)
#define sint2(a, b) ReadInt(a);ReadInt(b)
#define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c)
#define pint(a) WriteInt(a)
#define if_else(a, b, c) if (a) { b; } else { c; }
#define if_than(a, b) if (a) { b; }
#define test_pint1(a) printf("var1 = %d
", a)
#define test_pint2(a, b) printf("var1 = %d, var2 = %d
", a, b)
#define test_pint3(a, b, c) printf("var1 = %d, var2 = %d, var3 = %d
", a, b, c)

typedef double db;
typedef long long LL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef set<int> si;
typedef vector<int> vi;
typedef map<int, int> mii;

const int dx[8] = {0, 0, -1, 1};
const int dy[8] = {-1, 1, 0, 0};
const int maxn = 1e6 + 7;
const int maxm = 1e5 + 7;
const int maxv = 1e7 + 7;
const int max_val = 1e6 + 7;
const int MD = 998244353;
const int INF = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-10;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=0;while(isdigit(c)){x=x*10+c-'0';c=getchar();}}
template<class T>void WriteInt(T i) {int p=0;static int b[20];if(i == 0) b[p++] = 0;else while(i){b[p++]=i%10;i/=10;}for(int j=p-1;j>=0;j--)pchr('0'+b[j]);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

int pow_mod(int a, int b) {
    static int buf[100];
    int p = 0;
    while (b) {
        buf[p++] = b & 1;
        b >>= 1;
    }
    LL ans = 1;
    rep_down0(i, p) {
        ans = ans * ans % MD;
        if (buf[i]) ans = ans * a % MD;
    }
    return ans;
}

int cnt[maxn], c[maxn], f[maxn];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int T;
    cin >> T;
    while (T--) {
        mem0(cnt);
        mem0(c);
        int n, k;
        cin >> n >> k;
        int max_n = 0;
        rep_up0(i, n) {
            int x;
            sint(x);
            cnt[x]++;
            max_update(max_n, x);
        }
        rep_up1(i, max_n) {
            for (int j = i; j <= max_n; j += i) {
                c[i] += cnt[j];
            }
        }
        rep_up1(i, max_n) f[i] = (pow_mod(2, c[i]) + MD - 1) % MD;
        LL ans = 0;
        rep_down1(i, max_n) {
            for (int j = 2 * i; j <= max_n; j += i) {
                f[i] = (f[i] - f[j] + MD) % MD;
            }
            ans = (ans + (LL)f[i] * (pow_mod(i, k))) % MD;
        }
        cout << ans << endl;
    }
    return 0;
}