2019CCPC network game HD6707-- Du teach sieve

The meaning of problems

Request $ f (n, a, b) = \ sum_ {i = 1} ^ n \ sum_ {j = 1} ^ i gcd (i ^ aj ^ a, i ^ bj ^ b) [gcd (i, j) = 1] \% (10 ^ 9 + 7) $, $ 1 \ le n, a, b \ le 10 ^ 9 $, a total of $ T $ group test, only 10 sets of $ n-$ greater than $ 10 ^ 6 $.

analysis

First, when $ i, j $ coprime, $ a, b $ coprime with $ gcd (i ^ aj ^ a , you ^ bj ^ b) = ij $ ( proof see links ), can also play table guess Well a guess.

Can be introduced: $$ \ sum_ {d = 1} ^ {N} \ mu (d) \ cdot d \ sum_ {i = 1} ^ {\ lfloor \ frac {N} {d} \ rfloor} \ sum_ {j = 1} ^ {i} (ij) $$

Consider the latter part alone, $ \ sum_ {i = 1} ^ {k} \ sum_ {j = 1} ^ {i} (ij) = \ frac {k ^ 3-k} {6} $.

Then, only the left side of $ \ mu (d) \ cdot d $,

It with the identity function $ Id (n) = n $ Dirichlet convolution to give

$$\begin{align*}
(\mu(d)\cdot d)*Id(d)
& =  \sum_{d|n}(\mu(d)\cdot d)\cdot Id(\frac{n}{d})\\
& = \sum_{d|n}\mu(d) =  [n=1]
\end{align*}$$

The next set of Du teach sieved formula

$$\begin{align*}
S(n)
& = \sum\limits_{i=1}^n [i=1]-\sum\limits_{i=2}^ni\cdot S(\lfloor\dfrac{n}{i}\rfloor)\\
& = 1-\sum\limits_{i=2}^ni\cdot S(\lfloor\dfrac{n}{i}\rfloor)
\end{align*}$$

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

typedef long long ll;
const int maxn = 6e6 + 10;
const ll mod = 1e9+7;
const ll inv6 = 166666668;
int sum[maxn], mu[maxn], pri[maxn], pn;
bool vis[maxn];
map<int, int>mp_sum;
int n, a, b;

ll s2(ll i, ll j)
{
    return (i+j) * (j-i+1) / 2 % mod;
}

ll s3(ll k)
{
    return (k*k%mod - 1) * k % mod * inv6 % mod;
}

ll S(ll x)
{
    if(x < maxn)  return sum[x];
    if(mp_sum[x])  return mp_sum[x];
    ll ret = 1LL;
    for(int i = 2, j;i <= x;i = j+1)
    {
        j = x / (x / i);
        ret = (ret - s2(i, j) * (S(x/i))%mod) % mod;
    }
    return  mp_sum[x] = (ret + mod) % mod;
}

void pre()
{
    mu[1] = 1;
    for(int i = 2;i < maxn;i++)
    {
        if(!vis[i])
        {
            pri[++pn] = i;
            mu[i] = -1;
        }
        for(int j = 1;j <= pn && i * pri[j] < maxn; j++)
        {
            vis[i * pri[j]] = true;
            if(i % pri[j])  mu[i * pri[j]] = -mu[i];
            else
            {
                mu[i * pri[j]] = 0;
                break;
            }
        }
    }
    for(int i = 1;i < maxn;i++)  sum[i] = (sum[i-1] + i * mu[i]) % mod;
}

int main()
{
    pre();

    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%d", &n, &a, &b);
        ll ans  = 0;
        for(ll l = 1,r; l <= n;l = r+1)
        {
            r = n / (n / l);
            ans = (ans + (S(r) - S(l-1)) * s3(n/l)) % mod;
        }
        printf("%lld\n", (ans+mod)%mod);
    }
    return 0;
}

Initially opened MAXN = 2e6, will TLE; original blog open 6e6, and MLE, a long long int array into the job.

A standard is actually pushed into $ \ displaystyle ans = \ frac {\ sum _ {i = 1} ^ ni \ varphi (i) - 1} {2} $, less a divisible block.

However, through this solution, I deeply understand the temporal and spatial contradictions Du taught how to balance screen.

 

Reference Links: https://segmentfault.com/a/119000002017183