牛客练习赛69 F-解方程

传送门： https://ac.nowcoder.com/acm/contest/7329/F

题意

求解：$f_{i}mod998244353(1\le i\le n)$。

思路

$前置技能：\mu \ast I=\epsilon {\sigma }_q=i{d}_q\ast IF(n)={\sum }_{d|n}\mu (i)f(\frac{n}{d})$

$$\sum _{i|n}f(i){\sigma }_p(\frac{n}{i})={\sigma }_q(n)->(f\ast {\sigma }_p)n={\sigma }_q(n)$$

$先将\sigma 分解：$

$$(f\ast i{d}_p\ast I)n=(i{d}_q\ast I)n$$

$左右都乘上\mu :$

$$(f\ast i{d}_p)n=i{d}_q(n)$$

$$\sum _{i|n}f(i)\frac{n^p}{i^p}=n^q$$

$$\sum _{i|n}\frac{f(i)}{i^p}=n^{q-p}=i{d}_{q-p}(n)$$

$反演得：$

$$\frac{f(n)}{n^p}=\sum _{i|n}\frac{\mu (i)\ast n^{q-p}}{i^{q-p}}$$

$$\frac{f(n)}{n^q}=\sum _{i|n}\frac{\mu (i)}{i^{q-p}}$$

因为积性函数卷积性函数之后还是积性函数，所以能得到$\frac{f(n)}{n^q}$是积性函数。我们考虑与n互质的d，计算$f(d^k)$并且由莫比乌斯函数的性质得：

$$\frac{f(d^k)}{d^{kq}}=1-\frac{[k\ne 0]}{d^{q-p}}$$
$(根据上式，只有i=1或d时才\mu (i)满足，其他都是0)$

$所以再来看n，由基本算数定理得：n=p_1^{k_1}{p}_2^{k_2}...p_m^{k_m}，即：$

$$\frac{f(n)}{n^q}=\frac{f(p_1^{k_1})}{p_1^{q{k}_1}}\frac{f(p_2^{k_2})}{p_2^{q{k}_2}}...\frac{f(p_m^{k_m})}{p_m^{q{k}_m}}$$

$$\frac{f(n)}{n^q}=\prod _{d|n\&\&disaprime}(1-\frac{1}{d^{q-p}})$$

提前用欧拉筛处理$f(p^k)$即可，注意g数组的含义，因为当$i$时，$prime[j]$已经是$prime[j]\ast i$的因子的，而$f[i]$和$f[i\ast prime[j]]$两者之间的区别就是$prime[j{]}^q$，这还是比较难想到的。

Code(477MS)

#include <bits/stdc++.h>

using namespace std;

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

#define INF 0x7f7f7f
#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 P = 1e9 + 7;
// const double eps = 1e-6;
// const double pi = acos(-1);

const int N = 1e7 + 10;

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;
}

int prime[N];
bool is_prime[N];
int cnt;

ll q, p, n;
ll f[N], g[N];

void Eluer() {
    
    
    is_prime[1] = is_prime[0] = true;
    f[1] = 1;
    for(int i = 2;i < N; i++) {
    
    
        if(!is_prime[i]) {
    
    
            prime[cnt++] = i;
            g[i] = quick_pow(i, q);
            f[i] = (g[i] - quick_pow(i, p)) % mod;
        }
        for(int j = 0;j < cnt && prime[j] * i < N; j++) {
    
    
            is_prime[i * prime[j]] = true;
            if(i % prime[j] == 0) {
    
    
                f[i * prime[j]] = f[i] * g[prime[j]] % mod;
                break;
            }
            f[i * prime[j]] = f[i] * f[prime[j]] % mod;
        }
    }
}


void solve() {
    
    
    cin >> n >> p >> q;
    Eluer();
    ll ans = 0;
    for(int i = 1;i <= n; i++) {
    
    
        f[i] = (f[i] % mod + mod) % mod;
        ans ^= f[i];
    }
    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;
}