• 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
    ", 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 }
    View Code
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  • 原文地址:https://www.cnblogs.com/jklongint/p/4424901.html
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