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。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define lson l, m, rt << 1 26 #define rson m + 1, r, rt << 1 | 1 27 #define define_m int m = (l + r) >> 1 28 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 29 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 30 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 31 #define rep_down1(a, b) for (int a = b; a > 0; a--) 32 #define all(a) (a).begin(), (a).end() 33 #define lowbit(x) ((x) & (-(x))) 34 #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) {} 35 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 36 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 37 #define pchr(a) putchar(a) 38 #define pstr(a) printf("%s", a) 39 #define sstr(a) scanf("%s", a); 40 #define sint(a) ReadInt(a) 41 #define sint2(a, b) ReadInt(a);ReadInt(b) 42 #define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c) 43 #define pint(a) WriteInt(a) 44 #define if_else(a, b, c) if (a) { b; } else { c; } 45 #define if_than(a, b) if (a) { b; } 46 #define test_pint1(a) printf("var1 = %d ", a) 47 #define test_pint2(a, b) printf("var1 = %d, var2 = %d ", a, b) 48 #define test_pint3(a, b, c) printf("var1 = %d, var2 = %d, var3 = %d ", a, b, c) 49 50 typedef double db; 51 typedef long long LL; 52 typedef pair<int, int> pii; 53 typedef multiset<int> msi; 54 typedef set<int> si; 55 typedef vector<int> vi; 56 typedef map<int, int> mii; 57 58 const int dx[8] = {0, 0, -1, 1}; 59 const int dy[8] = {-1, 1, 0, 0}; 60 const int maxn = 1e6 + 7; 61 const int maxm = 1e5 + 7; 62 const int maxv = 1e7 + 7; 63 const int max_val = 1e6 + 7; 64 const int MD = 998244353; 65 const int INF = 1e9 + 7; 66 const double pi = acos(-1.0); 67 const double eps = 1e-10; 68 69 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 70 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();}} 71 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]);} 72 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 73 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 74 template<class T>T condition(bool f, T a, T b){return f?a:b;} 75 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 76 int make_id(int x, int y, int n) { return x * n + y; } 77 78 int pow_mod(int a, int b) { 79 static int buf[100]; 80 int p = 0; 81 while (b) { 82 buf[p++] = b & 1; 83 b >>= 1; 84 } 85 LL ans = 1; 86 rep_down0(i, p) { 87 ans = ans * ans % MD; 88 if (buf[i]) ans = ans * a % MD; 89 } 90 return ans; 91 } 92 93 int cnt[maxn], c[maxn], f[maxn]; 94 95 int main() { 96 //freopen("in.txt", "r", stdin); 97 //freopen("out.txt", "w", stdout); 98 int T; 99 cin >> T; 100 while (T--) { 101 mem0(cnt); 102 mem0(c); 103 int n, k; 104 cin >> n >> k; 105 int max_n = 0; 106 rep_up0(i, n) { 107 int x; 108 sint(x); 109 cnt[x]++; 110 max_update(max_n, x); 111 } 112 rep_up1(i, max_n) { 113 for (int j = i; j <= max_n; j += i) { 114 c[i] += cnt[j]; 115 } 116 } 117 rep_up1(i, max_n) f[i] = (pow_mod(2, c[i]) + MD - 1) % MD; 118 LL ans = 0; 119 rep_down1(i, max_n) { 120 for (int j = 2 * i; j <= max_n; j += i) { 121 f[i] = (f[i] - f[j] + MD) % MD; 122 } 123 ans = (ans + (LL)f[i] * (pow_mod(i, k))) % MD; 124 } 125 cout << ans << endl; 126 } 127 return 0; 128 }