计蒜客 Convex Hull

 

链接

$\sum \limits_{i=1}^n\sum\limits_{j=1}^i gay(i)=\sum \limits_{i=1}^ngay(i)(n-i+1)$

就是要求 $gay(n)$，$ngay(n)$ 的前缀和。把 $gay(n)$ 记成 $f(n)$

$$\begin{aligned} & \sum_{i=1}^n f(i)\\=&\sum_{i=1}^n i^2\mu^2(i)\\=& \sum_{i=1}^n i^2 \sum_{d^2|i}\mu(d)\\=& \sum_{d=1}^{\sqrt n} \mu(d) \sum_{i=1}^{n}i^2[d^2|i]\\=& \sum_{d=1}^{\sqrt n} \mu(d)d^4 \sum_{i=1}^{\lfloor \frac{n}{d^2} \rfloor}i^2 \end{aligned}$$

预处理出 $\mu(n)n^4$ 即可。

$nf(n)$ 的求法同上

#include <bits/stdc++.h>
#define ll long long

const int N = 1e5;
int prime[N + 7], prin, mu[N + 7];
bool vis[N + 7];
__int128 fs[N + 7], gs[N + 7];

void init() {
    mu[1] = 1;
    for (int i = 2; i <= N; i++) {
        if (!vis[i]) {
            prime[++prin] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= prin && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= N; i++) {
        fs[i] = fs[i - 1] + __int128(i) * i * i * i * mu[i];
        gs[i] = gs[i - 1] + __int128(i) * i * i * i * i * i * mu[i];
    }
}

ll MOD;

void M(ll &x) {
    if (x >= MOD) x -= MOD;
    if (x < 0) x += MOD;
}

ll sum2(ll n) {
    assert(MOD * 6 != 0);
    return __int128(n) * (n + 1) % (MOD * 6) * (n * 2 + 1) % (MOD * 6) / 6;
}

ll sum3(ll n) {
    assert(MOD * 2 != 0);
    ll temp = __int128(n) * (n + 1) % (MOD * 2) / 2;
    return __int128(temp) * temp % MOD;
}

int main() {
    init();
    ll n;
    while (~scanf("%lld%lld", &n, &MOD)) {
        ll ans, temp1 = 0, temp2 = 0;
        for (ll i = 1, j; i * i <= n; i = j + 1) {
            j = sqrt(n / (n / (i * i)));
            M(temp1 += __int128(fs[j] - fs[i - 1]) % MOD * sum2(n / (i * i)) % MOD);
            M(temp2 += __int128(gs[j] - gs[i - 1]) % MOD * sum3(n / (i * i)) % MOD);
        }
        M(ans = (__int128(n + 1) * temp1 % MOD - temp2) % MOD);
        printf("%lld\n", ans);
    }
    return 0;
}