ZOJ 3868 GCD expectation and BC39 HDU 5212 Code

The type of the two topics feel the same is the use of the idea of the repulsion to push forward and then go heavy.

HDU 5212 Test Instructions:

The number of n is calculated for each of the n numbers, and F (i), f (i) = GCD (A[i], a[j]) * (GCD (a[i], a[j])-1).

Consider the AI with AJ coprime when the value of f () is equal to 0 no need to calculate. Just calculate the number of all pairs of gcd with I as well (1<=i <=10000).

#include <algorithm> #include <cstdio> #include <vector> #include <cmath> #include <queue > #include <set> #include <map> #include <cstring> #include <cstdlib> #include <iostream >//#include <bits/stdc++.h> #define MAX 0x3f3f3f3f#define N 10005#define mod 10007#define Lson o<<1, L, m#d  Efine Rson o<<1|1, m+1, R#define Mem (a) memset (a, 0, sizeof (a)) typedef long Long ll;using namespace Std;const double    Pi = ACOs ( -1.0); int n, V[n], Dp[n], X;int main () {//freopen ("In.txt", "R", stdin);  while (CIN >> N) {mem (v);  MEM (DP);  int mx = 1;  for (int i = 0; i < n; i++) {scanf ("%d", &x);  v[x]++; mx = max (mx, x);} int ans = 0;for (int i = mx; i >= 1; i--) {Dp[i] = 0;int cnt = 0;for (int j = i; j <= mx; j + = i) {cnt + = V[j];if (J &G T i) {Dp[i] = (Dp[i]-dp[j])% mod + mod;dp[i]%= mod;} Dp[i] = (Dp[i] + cnt * CNT%mod)%mod;int mul = i * (i-1)%mod;ans + = mul * Dp[i];ans%= mod;} cout << ans << endl;} return 0;} 

Zhejiang University race GCD expectation

Test instructions is the K-square of the corresponding GCD for all subsets of the n number. The method is to enumerate 1 to the maximum value as GCD, how many multiples of I are counted as TMP, 2^tmp-1 must contain all the GCD as a subset of I but there may be gcd than I large, so to subtract I of twice times and more than the number of reverse can be counted out.

#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include < cstdio> #include <vector> #include <cmath> #include <queue> #include <set> #include <map > #define N 1000005#define mod 998244353#define MAX 0x3f3f3f3f#define Lson o<<1, L, M#define Rson o<<1|1, m+  1, R#define Mem (a) memset (A,0,sizeof (a)) typedef long Long ll;using namespace std;const double pi = ACOs ( -1.0); int n, V[n], X, MX, k;ll dp[n];ll POW (ll A, ll b) {ll res = 1;while (b) {if (b & 1) res = res * a%mod;b/= 2;a = a * a%mod;} return res;} ll solve () {ll res = 0;for (int i = mx; i > 0; i--) {Dp[i] = 0;ll cnt = 0;for (int j = i; j <= mx; j + = i) {cnt + = V[j ];if (J > i) dp[i] = ((Dp[i]-dp[j])%mod + mod)%mod; }dp[i] + = POW (2, CNT)-1;dp[i]%= mod;res = (res%mod + dp[i] * POW (i, k)%mod)%mod;} return res;}    int main () {//freopen ("In.txt", "R", stdin); int T;  Cin >> T;  while (t--) {cin >> n >> K;  mx = 0; MeM (v);  for (int i = 0; i < n; i++) {scanf ("%d", &x);  v[x]++; mx = max (mx, x);}    cout << Solve () << Endl;} return 0;}



ZOJ 3868 GCD expectation and BC39 HDU 5212 Code