Lonely Numbers

最新推荐文章于 2025-05-24 09:26:43 发布

千木流烟

最新推荐文章于 2025-05-24 09:26:43 发布

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文章标签：算法

本文链接：https://blog.youkuaiyun.com/2303_79062963/article/details/143352442

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I. Lonely Numbers

In number world, two different numbers are friends if they have a lot in common, but also each one has unique perks.

More precisely, two different numbers $a$ and $b$ are friends if $g c d (a, b)$ , $\frac{a}{gcd(a,b)}$ , $\frac{b}{gcd(a,b)}$ can form sides of a triangle.

Three numbers $a$ , $b$ and $c$ can form sides of a triangle if $a + b > c$ , $b + c > a$ and $c + a > b$ .

In a group of numbers, a number is lonely if it doesn’t have any friends in that group.

Given a group of numbers containing all numbers from $1, 2, 3, ..., n$ , how many numbers in that group are lonely?

Input

The first line contains a single integer $t$ $\leq t \leq 10^6)$ - number of test cases.

On next line there are $t$ numbers, $n_i$ $\leq n_i \leq 10^6)$ - meaning that in case $i$ you should solve for numbers $1, 2, 3, ..., n_i$

Output

For each test case, print the answer on separate lines: number of lonely numbers in group $1, 2, 3, ..., n_i$ .

Example

Input

3
1 5 10

Output

1
3
3

Note

For first test case, $1$ is the only number and therefore lonely.

For second test case where $n = 5$ , numbers $1$ , $3$ and $5$ are lonely.

For third test case where $n = 10$ , numbers $1$ , $5$ and $7$ are lonely.

code

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
 
using namespace std;


const int N = 1e6+10,INF=0x3f3f3f3f,mod=1e9+7;
 
typedef pair<int,int> PII;

int T=1;
int a[N];
bool st[N];
int b[N];
int c[N]; 
int d[N];
int sum[N];
int cnt;

void f(int n){
	for(int i=2;i<=n;i++)
    {
        if(!st[i]) a[cnt++] = i;
        if(a[cnt-1]*a[cnt-1]<=N) b[a[cnt-1]*a[cnt-1]]=1;
        c[a[cnt-1]]=1;
        for(int j=0;a[j]<=n/i;j++)
        {
            st[a[j]*i]=true;
            if(i%a[j]==0) break;
        }
    }
}

void solve(){
	int n;
    cin>>n;
    vector<int> t(n+5,0);
    int Max=-1;
    for(int i=0;i<n;i++) cin>>t[i],Max=max(Max,t[i]);
    f(Max);
//    cout<<Max<<endl;
	for(int i=1;i<=Max;i++) sum[i]=sum[i-1]+b[i];
	for(int i=1;i<=Max;i++){
		d[i]=d[i-1]+c[i];
//		cout<<"d["<<i<<"]:"<<d[i]<<endl; 
	}
    int ans=1;
	for(int i=0;i<n;i++){
		cout<<ans+d[t[i]]-sum[t[i]]<<endl;
	}
}

signed main(){
//	cin>>T; 
 	ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    while(T--){
        solve();
    }
    return 0;
}