题意

Sol

min25筛的板子题，直接筛出\(g(N, \infty)\)即可

筛的时候有很多trick，比如只存\(\frac{N}{x}\)的值，第二维可以滚动数组滚动掉

#include
    
     #define LL long long//#define int long long  using namespace std;const int MAXN = 2e6 + 10;int Lim, vis[MAXN], prime[MAXN], tot;LL N, g[MAXN], id[MAXN], cnt, pos1[MAXN], pos2[MAXN];void Get(int N) {    vis[1] = 1;    for(int i = 2; i <= N; i++) {        if(!vis[i]) prime[++tot] = i;        for(int j = 1; j <= tot && i * prime[j] <= N; j++) {            vis[i * prime[j]] = 1;            if(!(i % prime[j])) break;        }    }}LL get(LL x) {    return x <= Lim ? pos1[x] : pos2[N / x];}signed main() {    cin >> N; Lim = sqrt(N);    Get(Lim);    for(LL i = 1, j; i <= N; i = N / j + 1) {        j = N / i; id[++cnt] = j; g[cnt] = id[cnt] - 1;        j <= Lim ? pos1[j] = cnt : pos2[N / j] = cnt;;    }    for(int j = 1; j <= tot; j++)         for(LL i = 1; 1ll * prime[j] * prime[j] <= id[i]; i++)             g[i] -= g[get(id[i] / prime[j])] - (j - 1);    cout << g[1];    return 0;}