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\n", a)
 #define test_pint2(a, b) printf("var1 = %d, var2 = %d\n", a, b)
 #define test_pint3(a, b, c) printf("var1 = %d, var2 = %d, var3 = %d\n", 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[] = {, , -, };
 const int dy[] = {-, , , };
 const int maxn = 1e6 + ;
 const int maxm = 1e5 + ;
 const int maxv = 1e7 + ;
 const int max_val = 1e6 + ;
 const int MD = ;
 const int INF = 1e9 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=;while(isdigit(c)){x=x*+c-'';c=getchar();}}
 template<class T>void WriteInt(T i) {int p=;static int b[];if(i == ) b[p++] = ;else while(i){b[p++]=i%;i/=;}for(int j=p-;j>=;j--)pchr(''+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[];
     int p = ;
     while (b) {
         buf[p++] = b & ;
         b >>= ;
     }
     LL ans = ;
     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 = ;
         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(, c[i]) + MD - ) % MD;
         LL ans = ;
         rep_down1(i, max_n) {
             for (int j =  * 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 ;
 }