牛客练习赛76 F-phi and phi 莫比乌斯反演+差分

传送门： https://ac.nowcoder.com/acm/contest/10845/F

题意

$$求解ans[i]=\sum _{i=1}^n\sum _{j=1}^n\phi (ij)\phi [gcd(i,j)]$$

思路

$前置技能：\phi (ij)=\frac{\phi (i)\phi (j)gcd(i,j)}{\phi [gcd(i,j)]}$

$根据上式得：$
$$\sum _{i=1}^n\sum _{j=1}^n\phi (i)\phi (j)gcd(i,j)$$

$$\sum _{k=1}^{n}k\sum _{i=1}^n\sum _{j=1}^n\phi (i)\phi (j)[gcd(i,j)=k]$$

$$\sum _{k=1}^{n}k\sum _{i=1}^{\lfloor \frac{n}{k}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{k}\rfloor }\phi (ik)\phi (jk)[gcd(i,j)=1]$$

$$\sum _{k=1}^{n}k\sum _{i=1}^{\lfloor \frac{n}{k}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{k}\rfloor }\phi (ik)\phi (jk)\sum _{d|id|j}\mu (d)$$

$交换枚举顺序：$

$$\sum _{k=1}^{n}k\sum _{d=1}^{\lfloor \frac{n}{k}\rfloor }\mu (d)\sum _{i=1}^{\lfloor \frac{n}{kd}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{kd}\rfloor }\phi (ikd)\phi (jkd)$$

$$\sum _{k=1}^{n}k\sum _{d=1}^{\lfloor \frac{n}{k}\rfloor }\mu (d)[\sum _{j=1}^{\lfloor \frac{n}{kd}\rfloor }\phi (ikd){]}^2$$

$令T=kd并交换枚举顺序：$

$$\sum _{T=1}^n\sum _{k|T}k\mu (\frac{T}{k})[\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }\phi (iT){]}^2$$

$由\phi =id\ast \mu 得：$

$$\sum _{T=1}^n\phi (T)[\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }\phi (iT){]}^2$$

$设f(n/T)={\sum }_{j=1}^{\lfloor \frac{n}{T}\rfloor }\phi (iT)，则最终式为：$

$$ans[n]=\sum _{T=1}^n\phi (T)[f(n/T){]}^2$$

$但是这题要求的是ans[1-n]，我们发现$

$在枚举T时，n在[i\ast T,(i+1)\ast T)之间,\frac{n}{T}的值都一样，\phi (T)[f(n/T){]}^2的贡献不变。$

$所以我们在求ans[n]的过程中，对i\ast T做一个差分，最后在求一个前缀和即可。$

Code(445MS)

#pragma GCC optimize(2)
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pdd;

#define INF 0x3f3f3f3f
#define lowbit(x) x & (-x)
#define mem(a, b) memset(a , b , sizeof(a))
#define FOR(i, x, n) for(int i = x;i <= n; i++)

// const ll mod = 998244353;
 const ll mod = 1e9 + 7;
// const double eps = 1e-6;
// const double PI = acos(-1);
// const double R = 0.57721566490153286060651209;

const int N = 1e6 + 10;

int vis[N * 3], prime[N * 3], num;
ll phi[N * 3], ans[N * 3];

void init() {
     
    phi[1] = 1; num = 0;
    for (int i = 2; i < N; i++) {
     
        if (!vis[i]) {
     
            prime[++num] = i; phi[i] = i - 1;
        }
        for (int j = 1; j <= num && i * prime[j] < N; j++) {
     
            vis[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] * phi[prime[j]];
        }
    }
}

void solve() {
     
    init();
    int n; scanf("%d",&n);
    for(int T = 1;T <= n; T++) {
     
        ll sum = 0;
        for(int i = 1;i <= n / T; i++) {
     
            sum = (sum + phi[i * T]) % mod;
            ans[i * T] = (ans[i * T] + phi[T] * sum % mod * sum % mod + mod) % mod;
            ans[(i + 1) * T] = (ans[(i + 1) * T] - phi[T] * sum % mod * sum % mod + mod) % mod;
        }
    }
    for(int i = 1;i <= n; i++) {
     
        ans[i] = ((ans[i] + ans[i - 1]) % mod + mod) % mod;
        printf("%lld\n",ans[i]);
    }
    // cout << endl;
}

signed main() {
     
    solve();
}