[數論][Lucas定理] 51Nod 1778 小Q的集合

Solution$Solution$

通過枚舉⌊im⌋$\lfloor \frac{i}{m}\rfloor$ 和imodm$imodm$ 。由Lucas定理是可以獨立計算貢獻的。

ans====∑i=0n(ni)(ik−(n−i)k)2∑i=0n(⌊nm⌋⌊im⌋)(nmodmimodm)(ik−(n−i)k)2∑i=0⌊nm⌋(⌊nm⌋i)∑j=0nmodm(nmodmj)(ik−(n−i)k)22⌊nm⌋∑j=0nmodm(nmodmj)(ik−(n−i)k)2(57)(58)(59)(60)$$\begin{array}{}\text{(57)}& ans& =& \sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{i})(i^k-(n-i{)}^k{)}^2\text{(58)}& & =& \sum _{i=0}^n(\genfrac{}{}{0ex}{}{\lfloor \frac{n}{m}\rfloor }{\lfloor \frac{i}{m}\rfloor })(\genfrac{}{}{0ex}{}{nmodm}{imodm})(i^k-(n-i{)}^k{)}^2\text{(59)}& & =& \sum _{i=0}^{\lfloor \frac{n}{m}\rfloor }(\genfrac{}{}{0ex}{}{\lfloor \frac{n}{m}\rfloor }{i})\sum _{j=0}^{nmodm}(\genfrac{}{}{0ex}{}{nmodm}{j})(i^k-(n-i{)}^k{)}^2\text{(60)}& & =& 2^{\lfloor \frac{n}{m}\rfloor }\sum _{j=0}^{nmodm}(\genfrac{}{}{0ex}{}{nmodm}{j})(i^k-(n-i{)}^k{)}^2\end{array}$$

預處理ik$i^k$ 好像可以O(m)$\mathcal{O}(m)$ 做：
首先這是一個完全積性函數，可以線性篩，需要計算快速冪的只有質數，大概複雜度就是O(π(m)logm)$\mathcal{O}(\pi (m)\log m)$ ，就O(m)$\mathcal{O}(m)$ 了。

#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> Pairs;

const int N = 1010101;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}

int k, m, pcnt;
int fk[N], n[N];
int ndm, nmm, ans;
int s[N];
int fac[N], ifac[N];
int prime[N], vis[N];

inline int pwr(int a, int b) {
    int c = 1;
    while (b) {
        if (b & 1) c = (ll)c * a % m;
        b >>= 1; a = (ll)a * a % m;
    }
    return c;
}
inline void pre(int n) {
    ifac[1] = 1;
    for (int i = 2; i <= n; i++)
        ifac[i] = (ll)(m - m / i) * ifac[m % i] % m;
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= n; i++) {
        fac[i] = (ll)fac[i - 1] * i % m;
        ifac[i] = (ll)ifac[i - 1] * ifac[i] % m;
    }
    fk[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!vis[i]) {
            prime[++pcnt] = i;
            fk[i] = pwr(i, k);
        }
        for (int j = 1, x; j <= pcnt && (x = prime[j] * i) <= n; j++) {
            vis[x] = 1; fk[x] = (ll)fk[i] * fk[prime[j]] % m;
            if (i % prime[j] == 0) break;
        }
    }
}
inline int C(int n, int m) {
    return (ll)fac[n] * ifac[m] % ::m * ifac[n - m] % ::m;
}

int main(void) {
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    for (char c = get(); c >= '0' && c <= '9'; c = get())
        n[++*n] = c - '0';
    read(k); read(m); pre(m - 1);
    int cur = 0;
    for (int i = 1; i <= *n; i++) {
        cur = cur * 10 + n[i];
        nmm = (nmm * 10 + n[i]) % m;
        ndm = (ndm * 10 + cur / m) % (m - 1);
        cur %= m;
    }
    for (int j = 0; j <= nmm; j++)
        ans = (ans + (ll)pwr(fk[j] - fk[nmm - j] + m, 2) * C(nmm, j) % m) % m;
    ans = (ll)ans * pwr(2, ndm) % m;
    cout << ans << endl;
    return 0;
}