51Nod 1239 欧拉函数前n项和 杜教筛

 http://www.51nod.com/Challenge/Problem.html#!#problemId=1239

AC代码

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n");
#define debug(a,b) cout<<a<<" "<<b<<" ";
using namespace std;
const int maxn=1e6+10,inf=0x3f3f3f3f;
typedef long long ll;
const ll mod = 1000000007;
typedef pair<int,int> pii;
int check[maxn],prime[maxn],phi[maxn],sum[maxn];
void Phi(int N)//莫比乌斯函数线性筛
{
    int pos=0;sum[1]=phi[1]=1;
    for(int i = 2 ; i <= N ; i++)
    {
        if (!check[i])
            prime[pos++] = i,phi[i]=i-1;
        for (int j = 0 ; j < pos && i*prime[j] <= N ; j++)
        {
            check[i*prime[j]] = 1;
            if (i % prime[j] == 0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
        sum[i]=(sum[i-1]+phi[i])%mod;
    }
}
unordered_map<ll,ll> ma;
ll inv2=500000004;
ll solve(ll n)
{
    if(n<=1e6)
        return sum[n];
    else if(ma.count(n))
        return ma[n];
    ll temp = ((n%mod)*((n+1)%mod)%mod)*inv2%mod;
    for(ll i=2,j;i<=n;i=j+1)
    {
        j=n/(n/i);
        temp = (temp-solve(n/i)*(j-i+1)%mod+mod)%mod;
    }
    return ma[n]=temp;
}
int main()
{
    ll n;
    Phi(1e6);
    scanf("%lld",&n);
    printf("%lld\n",solve(n));
}