HDU6954 2021多校 Minimum spanning tree (字首和)
阿新 • • 發佈:2021-07-21
#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; }