bzoj3930 CQOI2015 选数

题面：bzoj3930

题意：从$[L,H]$选出$n$个数，使它们的最大公约数为$k$的方案数，对$1e9+7$取模。

$1\le n,k,L,H\le 10^9$，$H-L\le 10^5$

题解：即求${\sum }_{a_1=L}^H{\sum }_{a_2=L}^H\dots {\sum }_{a_n=L}^H[gc{d}_{i=1}^n(a_i)=k]$

令$f(x)={\sum }_{a_1=L}^H{\sum }_{a_2=L}^H\dots {\sum }_{a_n=L}^H[gc{d}_{i=1}^n(a_i)=x]$

$g(x)={\sum }_{x\mid d}f(d)=(\lfloor \frac{H}{x}\rfloor -\lfloor \frac{L-1}{x}\rfloor {)}^n$

莫比乌斯反演：$f(x)={\sum }_{x\mid d}\mu (\frac{d}{x})g(d)={\sum }_{x\mid d}\mu (\frac{d}{x})(\lfloor \frac{H}{d}\rfloor -\lfloor \frac{L-1}{d}\rfloor {)}^n$

答案即为$f(k)={\sum }_{k\mid d}\mu (\frac{d}{k})(\lfloor \frac{H}{d}\rfloor -\lfloor \frac{L-1}{d}\rfloor {)}^n={\sum }_{i=1}^{\lfloor \frac{H}{k}\rfloor }\mu (i)(\lfloor \frac{H}{ki}\rfloor -\lfloor \frac{L-1}{ki}\rfloor {)}^n$

明显可以数论分块。但是$\lfloor \frac{H}{k}\rfloor$可能达到$1e9$，需要快速对${\sum }_{i=1}^n\mu (i)$求和。

考虑杜教筛：

令$S(n)={\sum }_{i=1}^n\mu (i)$

$g(1)S(n)={\sum }_{i=1}^n(g\ast \mu )(i)-{\sum }_{i=2}^{n}g(i)S(\frac{n}{i})$

令$g(i)=1$

$S(n)=1-{\sum }_{i=2}^{n}S(\frac{n}{i})$

记忆化算出$S(n)$

#include 
using namespace std;
typedef long long ll ;
const int maxn = 1e7 + 10, mod = 1e9 + 7 ;
const ll INF = 1e18 ;
int prime[maxn], tot ;
ll mu[maxn] ;
bool vis[maxn] ;
map<ll, ll> val ;
void init () {
    mu[1] = 1 ;
    for (int i = 2; i < maxn; i ++) {
        if (!vis[i]) prime[++ tot] = i, mu[i] = -1 ;
        for (int j = 1; j <= tot && i * prime[j] < maxn; j ++) {
            vis[i * prime[j]] = 1 ;
            if (i % prime[j] == 0) {
                mu[i * prime[j]] = 0; break ;
            }
            mu[i * prime[j]] = -mu[i] ;
        }
    }
    for (int i = 1; i < maxn; i ++) mu[i] = (mu[i - 1] + mu[i]) % mod ;
}
inline ll power (ll x, ll y) {
    ll res = 1 ;
    while (y) {
        if (y & 1) res = res * x % mod ;
        x = x * x % mod; y >>= 1 ;
    }
    return res ;
}
ll cal (ll n) {
    if (n < maxn) return mu[n] ;
    if (val[n]) return val[n] ;
    ll res = 1 ;
    for (ll i = 2, nxt = 0; i <= n; i = nxt + 1) {
        nxt = n / (n / i) ;
        res = (res - (nxt - i + 1) * cal (n / i) % mod) % mod ;
    }
    return val[n] = (res + mod) % mod ;
}
int main() {
    ll n, k, l, h ;
    cin >> n >> k >> l >> h ;
    init () ;
    -- l /= k; h /= k ;
    ll ans = 0 ;
    for (ll i = 1, nxt = 0; i <= h; i = nxt + 1) {
        nxt = min (l / i ? l / (l / i) : INF, h / (h / i)) ;
        ans = (ans + power (h / i - l / i, n) * (cal (nxt) - cal (i - 1)) % mod) % mod ;
    }
    cout << (ans + mod) % mod << endl ;
    return 0 ;
}