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。

  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\n", a)

 47 #define test_pint2(a, b) printf("var1 = %d, var2 = %d\n", a, b)

 48 #define test_pint3(a, b, c) printf("var1 = %d, var2 = %d, var3 = %d\n", 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 }

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