[CQOI2015]选数

推式子

根据题意可写出式子：
$$\begin{gathered} \sum _{a_1=L}^H\sum _{a_2=L}^H\cdots \sum _{a_n=L}^H[gcd(a_1,a_2\dots a_n)=k] \\ \sum _{a_1=\lceil \frac{L}{k}\rceil }^{\lfloor \frac{H}{k}\rfloor }\sum _{a_2=\lceil \frac{L}{k}\rceil }^{\lfloor \frac{H}{k}\rfloor }\cdots \sum _{a_n=\lceil \frac{L}{k}\rceil }^{\lfloor \frac{H}{k}\rfloor }[gcd(a_1,a_2\dots a_n)=k] \\ \sum _{k=1}^{\lfloor \frac{H}{k}\rfloor }\mu (k){\left(\lfloor \frac{H}{kd}\rfloor -\lceil \frac{L}{kd}\rceil +1\right)}^n \\ 提前处理一下左右端点 \\ \sum _{k=1}^{\lfloor \frac{H}{k}\rfloor }\mu (k){\left(\lfloor \frac{H}{kd}\rfloor -\lfloor \frac{L-1}{kd}\rfloor \right)}^n \end{gathered}$$
非常简单的反演公式，所以我们只要按照这个公式套上杜教筛就行了。

代码

/*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 = 1e6 + 10, mod = 1e9 + 7;int prime[N], mu[N], cnt;
ll n, k, L, H;bool st[N];void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {mu[i] += mu[i - 1];}
}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;
}map<ll, ll> ans_s;ll S(ll n) {if(n < N) return mu[n];if(ans_s.count(n)) return ans_s[n];ll ans = 1;for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = ((ans - (r - l + 1) * S(n / l) % mod) % mod + mod) % mod;}return ans_s[n] = ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();n = read(), k = read(), L = read(), H = read();ll ans = 0;L = (L - 1) / k, H = H / k;for(ll l = 1, r; l <= H; l = r + 1) {r = min( L / l ? L / (L / l) : inf , H / (H / l));ans = (ans + 1ll * (((S(r) - S(l - 1)) % mod + mod) % mod) * quick_pow((H / l) - (L / l), n) % mod) % mod;}cout << ans << endl;return 0;
}