HDU 6750 Function（莫比乌斯反演）（2020百度之星初赛1）

Function

推式子

$$\begin{gathered} S(n)=\sum _{i=1}^n\sum _{d\mid i}d[gcd(d,\frac{i}{d})==1] \\ =\sum _{d=1}^{n}d\sum _{d\mid i}[gcd(d,\frac{i}{d})==1] \\ =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}[gcd(d,i)==1] \\ =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{k\mid gcd(d,i)}\mu (k) \\ =\sum _{k=1}^n\mu (k)k\sum _{d=1}^{\frac{n}{k}}d\sum _{i=1}^{\frac{n}{k^{2}d}} \\ t=k^{2}d \\ =\sum _{t=1}^n\frac{n}{t}\sum _{k^2\mid t}\mu (k)k\frac{t}{k^2} \\ =\sum _{k=1}^{\sqrt{n}}\mu (k)k\sum _{k^2\mid t}\frac{n}{t}\frac{t}{k^2} \\ i=\frac{t}{k^2} \\ =\sum _{k=1}^{\sqrt{n}}\mu (k)k\sum _{i=1}^{\frac{n}{k^2}}\frac{n}{i{k}^2}i \end{gathered}$$

​

代码

注意随手取模，long long $\times$long long会溢出！！！

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int inf = 0x3f3f3f3f;
const double eps = 1e-7;

const int N = 1e6 + 10, mod = 1e9 + 7, inv2 = mod + 1 >> 1;

int prime[N], cnt;

ll mu[N];

bool st[N];

void init() {
    mu[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < cnt && i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1; i < N; i++) {
        mu[i] = (mu[i - 1] + 1ll * i * mu[i] % mod + mod) % mod;
    }
}
ll calc1(ll l, ll r) {
    return 1ll * (l + r) % mod * ((r - l + 1) % mod) % mod * inv2 % mod;
}

ll calc2(ll n) {
    ll ans = 0;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = n / (n / l);
        ans = (ans + 1ll * (n / l) % mod * calc1(l, r) % mod) % mod;
    }
    return ans;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    init();
    int T;
    scanf("%d", &T);
    while(T--) {
        ll n, ans = 0;
        scanf("%lld", &n);
        int m = sqrt(n);
        for(ll l = 1, r; l <= m; l = r + 1) {
            r = min((int)sqrt(n / (n / (l * l))), m);
            ans = (ans + 1ll * ((mu[r] - mu[l - 1]) % mod + mod) % mod * calc2(n / (l * l)) % mod) % mod;
        }
        printf("%lldn", ans);
    }
	return 0;
}

最后

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