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 }
时间： 2024-08-07 01:56:20