bzoj4804: 欧拉心算 欧拉筛

Posted 2021-01-01 acjiumeng

题意：求(sum_{i=1}^nsum_{j=1}^nphi(gcd(i,j)))
题解：(sum_{i==1}^nsum_{j=1}^nsum_{d=1}^n[gcd(i,j)==d]*phi(d))
(=sum_{d=1}^nphi(d)*sum_{i=1}^nsum_{j=1}^n[gcd(i,j)==d])
(=sum_{d=1}^nphi(d)*sum_{i=1}^{lfloor frac{n}{d} floor}[gcd(i,j)==1])
(=sum_{d=1}^nphi(d)*(2*sum_{i=1}^{lfloor frac{n}{d} floor}phi({lfloor frac{n}{d} floor})-1))
(=sum_{d=1}^nphi(d)*sum(lfloor frac{n}{d} floor)-sum(n))
求个前缀和分块搞一搞就好了

/**************************************************************
    Problem: 4804
    User: walfy
    Language: C++
    Result: Accepted
    Time:4108 ms
    Memory:167304 kb
****************************************************************/
 
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
 
using namespace std;
 
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=10000000+10,maxn=400000+10,inf=0x3f3f3f3f;
 
int prime[N],cnt,phi[N];
bool mark[N];
ll sum[N];
void init()
{
    phi[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!mark[i]){prime[++cnt]=i,phi[i]=i-1;}
        for(int j=1;j<=cnt&&i*prime[j]<N;j++)
        {
            mark[i*prime[j]]=1;
            phi[i*prime[j]]=phi[i]*phi[prime[j]];
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
        }
    }
    for(int i=1;i<N;i++)sum[i]=sum[i-1]+phi[i];
}
int main()
{
    init();
    int T;scanf("%d",&T);
    while(T--)
    {
        ll n,ans=0;
        scanf("%lld",&n);
        for(ll i=1,j;i<=n;i=j+1)
        {
            j=n/(n/i);
            ans+=(sum[j]-sum[i-1])*sum[n/i];
        }
        printf("%lld
",2*ans-sum[n]);
    }
    return 0;
}
/********************
 
********************/