51nod 237 最大公约数之和 V3 杜教筛

Code:

#include <bits/stdc++.h>
#include <tr1/unordered_map>

#define setIO(s) freopen(s".in","r",stdin)
#define ll long long 
#define ull unsigned long long 
#define maxn 10000000
#define mod 1000000007 
#define inv 500000004
 
using namespace std;
using namespace tr1;
int vis[maxn],prime[maxn],tot; 
ll phi[maxn]; 
unordered_map<ll,ull>ansphi;
void init(){
    phi[1] = 1; 
    for(int i=2;i<maxn; ++i) {
        if(!vis[i]) prime[++tot]=i,phi[i] = i-1; 
        for(int j=1;j<=tot&&i*prime[j]<maxn;++j) {
            vis[i*prime[j]]=1;
            if(i%prime[j]!=0) phi[i*prime[j]]=phi[i]*(prime[j]-1); 
            else {
                phi[i*prime[j]]=phi[i]*(prime[j]);
                break; 
            }
        }
    }    
    for(int i=1;i<maxn;++i) phi[i]+=phi[i-1],phi[i]%=mod; 
}
ll solve(ll n){
    if(n < maxn) return phi[n];
    if(ansphi[n]) return ansphi[n];
    ll ans=(ull)(((n%mod)*((n+1)%mod) %mod)*(inv%mod))%mod; 
    ll ans2=0; 
    for(ll l=2,r;l<=n;l=r+1) {
        r=n/(n/l);  
        ans2+=(ll)(r-l+1)*solve(n/l);  
        ans2%=mod; 
    }
    return ansphi[n]=(ans+mod-ans2)%mod; 
}
int main(){
    //setIO("input"); 
    init(); 
    ll n,ans=0,ans1,tmp;            
    scanf("%lld",&n);
    for(ll l=1,r;l<=n;l=r+1){                        
        r=(n/(n/l)); 
        ans+=(((n/l)%mod)*((n/l)%mod)%mod*(solve(r)+mod-solve(l-1))%mod)%mod;  
        ans%=mod;    
    }
    printf("%lld",ans); 
    return 0; 
}

  

原文地址:https://www.cnblogs.com/guangheli/p/10542261.html