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Sample input

11 10

Sample Output

10
5
7
5
8
3
9
5
7
4
10

 

 (Hee hee hee, only to get help several dalao know, you must be in memory of)

This question is the use of inclusion and exclusion ah

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 100005;
int n, prime[N], tot, cnt;
bool vis[N];
ll m, num[N], ans;

void getprime() {
    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]] = true;
            if (i % prime[j] == 0)
                break;
        }
    }
}

void work(int n) {
    for (int i = 1; i <= tot && prime[i] <= n; i++) {
        int x = prime[i];
        if (n % x == 0) {
            num[++cnt] = x;// 
            while (n % x == 0)
                n /= x;
        }
    }
} 

// lose its not relatively prime, the rest is prime to the number of friends 
 // not prime to gnaw cut it, looking for his prime factors of myself, equally sure that he's not prime factors of prime to friends 
void DFS ( int A, int p, int S) { // number p approximately, s is a positive or negative (plus or minus) 
    IF (! A) { 
        ANS + = (LL) * S m / p; // m / p m is the number of a multiple of p in 
        return ; 
    } 
    // see the number of prime factors, a -> plus, two -> Save, three -> plus ... 
    DFS (a - . 1 , p, S ); 
    DFS (A - 1 , * P NUM [A], - 1 * S);
     / * For example, there factor 3,4,5 
                     1
                  / \  
                / \
               . 3. 1 
             / \ / \ 
           . 3 *. 4. 3. 4. 1 
          / \ / \ / \ / \ 
      . 3 *. 4 *. 5. 3 *. 4. 3 *. 5. 3. 4 *. 5. 4. 5. 1 
    * / 
} 

Signed main () { 
    CIN > > >> n- m; 
    getprime (); // prime table 
    for ( int I = . 1 ; I <= n-; I ++ ) { 
        ANS = 0 ; 
        CNT = 0 ; // quality factor number 
        Work (I); // seeking the number of prime factors of i 
        DFS (CNT, . 1 , . 1 );
        printf("%lld\n", ans);
    }
}