HDU6954 2021多校 Minimum spanning tree （前缀和）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}
 
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const int maxn=10000000+10;
const double eps=1e-7;
 
ll sum[maxn],vis[maxn],pri[maxn],idx;
 
void init(){
	for(int i=2;i<maxn;i++){
		if(!vis[i]) pri[++idx]=i;
		for(int j=1;j<=idx&&i*pri[j]<maxn;j++){
			vis[i*pri[j]]=1;
			if(i%pri[j]==0) break;
		}
		if(i==2) continue;
		if(vis[i]) sum[i]=sum[i-1]+i;
		else sum[i]=sum[i-1]+i*2;
	}
}
 
int main(){
	init();
    int _=read;
    while(_--){
    	int n=read;
    	printf("%lld\n",sum[n]);
    }
    return 0;
}