51nod 1238 杜教篩

傳送門：51nod 1238

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題意

求

G(N)=∑i=1N∑j=1Nlcm(i,j)

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題解

首先

G(N)=∑i=1N∑j=1Nlcm(i,j)=2∑i=1N∑j=1ilcm(i,j)−∑i=1Nlcm(i,i)=2∑i=1Ni∑d|i∑u=1idu[gcd(u,id)=1]−N(N+1)2=2∑i=1Ni∑d|iidϕ(id)+[id=1]2−N(N+1)2=∑i=1Ni∑d|iidϕ(id)+[id=1]−N(N+1)2=∑i=1Ni(1+∑d|iidϕ(id))−N(N+1)2=∑i=1Ni∑d|iidϕ(id)=∑i=1Ni∑d|idϕ(d)

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其次枚舉因子對i的貢獻:

G(N)=∑i=1Ni∑d|idϕ(d)=∑i=1N∑d=1⌊Ni⌋dϕ(d)

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可用分塊:
令

F(x)=∑i=1xiϕ(i)

預處理N23 以內的F(i)
對於x>N23
杜教篩求解:

F(x)=∑i=1xiϕ(i)=∑i=1xi(i−∑d|i,d<iϕ(d))=∑i=1xi2−∑i=1xi∑d|i,d<iϕ(d))=x(2x+1)(x+1)6−∑i=2xi∑d=1⌊xi⌋dϕ(d)=x(2x+1)(x+1)6−∑i=2xiF(⌊xi⌋)

分塊求解：

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code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll mod = 1e9 + 7;
const ll inv = (mod + 1) / 2;
const ll inv2 = inv * (mod + 1) / 3 % mod;
const int mo = 2333333;
const int N = 1000001;

bool isPrime[N];
int prime[N];
ll phi[N];
ll sum[N];

void init()
{
    int cnt = 0;
    memset(isPrime, true, sizeof isPrime);
    isPrime[1] = false;
    phi[1] = 1;
    sum[1] = 1;
    for(int i = 2; i < N; ++i)
    {
        if(isPrime[i])
        {
            prime[++cnt] = i;
            phi[i] = i - 1;
        }
        for(int j = 1; j <= cnt && i * prime[j] < N; ++j)
        {
            isPrime[i * prime[j]] = false;
            if(i % prime[j] == 0)
            {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
    }
    for(ll i = 2; i < N; ++i)
    {
        sum[i] = (sum[i - 1] + i * i * phi[i] % mod) % mod;
    }
}


ll Sum(ll x)
{
    return x % mod * ((x * 2 + 1) % mod) % mod * ((x + 1) % mod) % mod * inv2 % mod;
}
ll getSum(ll R, ll L)
{
    return ((Sum(R) - Sum(L)) % mod + mod) % mod;
}
ll n;

int next[mo], last[mo];
ll t[mo], v[mo];
int l;
void add(int x, ll y, ll z)
{
    t[++l] = y;
    next[l] = last[x];
    last[x] = l;
    v[l] = z;
}
ll cal(ll x)
{
    if(x < N) return sum[x];
    int k = x % mo;
    for(int i = last[k]; i; i = next[i])
        if(t[i] == x) return v[i];
    ll res = x % mod * ((x + 1) % mod) % mod * inv % mod;
    res = res * res % mod;
   // if(x == 10000000000) cout << res << endl;
    ll r;
    for(ll i = 2; i <= x; i = r + 1)
    {
        ll tmp = x / i;
        r = x / tmp;
        ll mul = getSum(r, i - 1);
        //if(x == 10000000000)cout << mul << " ";
        ll ans = cal(tmp) * mul % mod;
       // if(x == 10000000000)cout << ans << " ";
        res = ((res - ans) % mod + mod) % mod;
       // if(x == 10000000000) cout << res << endl;
    }
    add(k, x, res);
    return res;
}


ll Calc(ll x)
{
    ll res = 0;
    ll r;
    for(ll i = 1; i <= x; i = r + 1)
    {
        ll tmp = x / i;
        r = x / tmp;
        ll mul = ((r - i + 1) % mod) % mod * ((r + i) % mod) % mod * inv % mod;
       // cout << mul << " ";
        res = (res + mul * cal(tmp)  % mod) % mod;
        //cout << res << endl;
    }
    r = 0;
  /**  r = x * inv % mod;
    ll mul = 1;
    for(ll i = 1; i * i <= x; ++i)
    {
        if(x % i == 0)
        {
            mul = (mul + i * getPhi(i)) % mod;
            if(i * i != x)  mul = (mul + x / i * getPhi(x / i)) % mod;
        }
    }
    r = r * mul % mod;*/
    return ((res - r) % mod + mod) % mod;
}
int main()
{

    init();
    cin >> n;
 // cout << inv << " " <<inv2 << endl;
    cout << Calc(n) << endl;

    return 0;
}