【2020杭电多校第六场】 6833 Very Easy Math Problem

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6833

借鉴大佬： https://blog.csdn.net/weixin_44282912/article/details/107844614 https://blog.csdn.net/oampamp1/article/details/107865300

题意

$$求解\sum_{a_1=1}^n\sum_{a_2=1}^n…\sum_{a_x=1}^n\left(\prod_{j=1}^xa_j^k\right)f(gcd(a_1,a_2…a_x))gcd(a_1,a_2…a_x)$$

思路

$令d=gcd(a_1,a_2…a_x)得：$

$$\sum_{a_1=1}^n\sum_{a_2=1}^n…\sum_{a_x=1}^n\left(\prod_{j=1}^xa_j^k\right)f(d)d$$

$枚举d得：$

$$\sum_{d=1}^ndf(d)\sum_{a_1=1}^n\sum_{a_2=1}^n…\sum_{a_x=1}^n\left(\prod_{j=1}^xa_j^k\right)[gcd(a_1,a_2…a_x)=d]$$

$$\sum_{d=1}^nd^{kx+1}f(d)\sum_{a_1=1}^{\left \lfloor \frac{n}{d} \right \rfloor }\sum_{a_2=1}^{\left \lfloor \frac{n}{d} \right \rfloor }…\sum_{a_x=1}^{\left \lfloor \frac{n}{d} \right \rfloor }\left(\prod_{j=1}^xa_j^k\right)[gcd(a_1,a_2…a_x)=1]$$

$$\sum_{d=1}^nd^{kx+1}f(d)\sum_{a_1=1}^{\left \lfloor \frac{n}{d} \right \rfloor }\sum_{a_2=1}^{\left \lfloor \frac{n}{d} \right \rfloor }…\sum_{a_x=1}^{\left \lfloor \frac{n}{d} \right \rfloor }\left(\prod_{j=1}^xa_j^k\right)\sum_{ta_1\;ta2\;…\;ta_x}\mu(t)$$

$$\sum_{d=1}^nd^{kx+1}f(d)\left (\sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor }i^k\right)^x\sum_{ta_1\;ta2\;…\;ta_x}\mu(t)$$

$枚举t得：$

$$\sum_{d=1}^nd^{kx+1}f(d)\sum_{t=1}^{\left \lfloor \frac{n}{d} \right \rfloor }\mu(t)\left (\sum_{i=1}^{\left \lfloor \frac{n}{dt} \right \rfloor }(it)^k\right)^x$$

$$\sum_{d=1}^nd^{kx+1}f(d)\sum_{t=1}^{\left \lfloor \frac{n}{d} \right \rfloor }\mu(t)t^{kx}\left (\sum_{i=1}^{\left \lfloor \frac{n}{dt} \right \rfloor }i^k\right)^x$$

$令T=dt并枚举T得：$

$$\sum_{T=1}^n\left (\sum_{i=1}^{\left \lfloor \frac{n}{dt} \right \rfloor }i^k\right)^xT^{kx}\sum_{dT}df(d)\mu(\frac{T}{d})$$

$令g(T)=\sum_{dT}df(d)\mu(\frac{T}{d})，则$

$$\sum_{T=1}^n\left (\sum_{i=1}^{\left \lfloor \frac{n}{T} \right \rfloor }i^k\right)^xT^{kx}g(T)$$

$O(nlogn)筛f和g，并对T^{kx}g(T)和\sum_{i=1}^{\left \lfloor \frac{n}{T} \right \rfloor }i^k做前缀和，最后\sqrt n分块计算，复杂度为O(nlogn+T\sqrt n)$

Code（2121MS）

##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 int N = 2e5 + 10;

ll T, k, x;

int mu[N]; // 莫比乌斯函数
bool is_prime[N];
int prime[N];
int cnt;
ll g[N], f[N];
ll sumG[N], sumi[N];

ll quick_pow(ll a, ll b) {
    ll ans = 1;
    while(b) {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans % mod;
}

void Mobi() // 莫比乌斯函数初始化
{
    f[1] = mu[1] = 1;
    is_prime[0] = is_prime[1] = true;
    for(int i = 2;i < N; i++) {
        f[i] = 1;
        if (!is_prime[i]) {
            mu[i] = -1;
            prime[++cnt] = i;
        }
        for (int j = 1; j <= cnt && i * prime[j] < N; j++) {
            is_prime[i * prime[j]] = true;
            if (i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
    // 筛f函数
    for(int d = 2;d * d < N; d++) {
        for(int i = d * d;i < N; i += d * d) {
            f[i] = 0;
        }
    }
    // 筛g函数
    for(int d = 1;d < N; d++) {
        for(int i = d;i < N; i += d) {
            g[i] = (g[i] + 1ll * d * f[d] % mod * mu[i / d] % mod + mod) % mod;
        }
    }
    // 处理sumG函数和sumi函数前缀和
    for(int i = 1;i < N; i++) {
        ll t = quick_pow(i, k);
        sumi[i] = (sumi[i - 1] + t) % mod;
        sumG[i] = (sumG[i - 1] + quick_pow(t, x) * g[i] % mod) % mod;
    }
}

void solve() {
    cin >> T >> k >> x;
    Mobi();
    while(T--) {
        int n;
        cin >> n;
        ll ans = 0;
        for(int l = 1, r;l <= n; l = r + 1) {
            r = min(n, n / (n / l));
            ans = (ans + (sumG[r] - sumG[l - 1] + mod) % mod * quick_pow(sumi[n / l], x) % mod) % mod;
        }

        cout << ans << endl;
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    //cin.tie(nullptr);
    //cout.tie(nullptr);
##ifdef FZT_ACM_LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
##else
    solve();
##endif
    return 0;
}

本文作者：jujimeizuo
本文地址： https://blog.jujimeizuo.cn/2020/10/07/hdu-6833/
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