P6156 简单题

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n(i+j{)}^{k}f(gcd(i,j))gcd(i,j) \\ =\sum _{i=1}^n\sum _{j=1}^n(i+j{)}^k{\mu }^2(gcd(i,j))gcd(i,j) \\ =\sum _{d=1}^n{\mu }^2(d)d^{k+1}\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}(i+j{)}^k(gcd(i,j)==1) \\ =\sum _{d=1}^n{\mu }^2(d)d^{k+1}\sum _{p=1}^{\frac{n}{d}}{p}^2\mu (p)\sum _{i=1}^{\frac{n}{pd}}\sum _{j=1}^{\frac{n}{pd}}(i+j{)}^k \\ t=pd,f(n)=\sum _{i=1}^n\sum _{j=1}^n(i+j{)}^k \\ =\sum _{t=1}^{n}f(\frac{n}{t})t^k\sum _{d\mid t}{\mu }^2(d)\mu (\frac{t}{d})d \end{gathered}$$

我们先对$f(n)$分析：
$$\begin{gathered} 假定a(n)=\sum _{i=1}^n{i}^k，b(n)=\sum _{i=1}^n\sum _{j=1}^i{j}^k=\sum _{i=1}^{n}a(i) \\ 简单化简一下就可以得到f(n)=b(2n)-2b(n) \\ 显然我们可以通过欧拉筛得到i^k，然后一次前缀和得到a(i),再一次前缀和得到b(i) \end{gathered}$$
接下来我们对$f(n)={\sum }_{d\mid n}{\mu }^2(d)\mu (\frac{n}{d})d$分析：
$$\begin{gathered} 显然这是一个积性函数，所以f(n)=f(p^k)f(\frac{n}{p^k})， \\ 当k=0,f(1)=1, \\ 当k=1,f(p)=p-1, \\ 当k=2,f(p^2)=-p, \\ 当k>=3，显然对任意一项d，或者\frac{n}{d}一定有一项存在平方项，所以\mu 会变成0, \\ 所以k>=3f(p^k)=0 \\ 所以这个积性函数我们只要在欧拉筛中特判处理一下即可。 \end{gathered}$$

代码

P6222 「P6156 简单题」加强版

这道题目的加强版本，由于我这里采用的是线性复杂度的算法，所以只要稍加修改即可过这道题。

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e7 + 10, mod = 998244353;int prime[N], cnt;ll f[N], pw[N], n, k;bool st[N];ll quick_pow(ll a, ll n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}void init() {f[1] = pw[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;f[i] = i - 1;pw[i] = quick_pow(i, k);}for(int j = 0; j < cnt && i * prime[j] < N; j++) {pw[i * prime[j]] = pw[i] * pw[prime[j]] % mod;st[i * prime[j]] = 1;if(i % prime[j] == 0) {int temp = i / prime[j];if(temp % prime[j]) f[i * prime[j]] = (mod - 1ll * prime[j] * f[temp] % mod) % mod;break;}f[i * prime[j]] = f[i] * f[prime[j]] % mod;}}for(int i = 1; i < N; i++) f[i] = (f[i - 1] + f[i] * pw[i] % mod) % mod, pw[i] = (pw[i] + pw[i - 1]) % mod;for(int i = 1; i < N; i++) pw[i] = (pw[i] + pw[i - 1]) % mod;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(), k = read(), init();ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + (f[r] - f[l - 1] + mod) % mod * (((pw[2 * (n / l)] - 2 * pw[n / l]) % mod + mod) % mod) % mod) % mod;}printf("%lld\n", ans);return 0;
}