[CQOI2015]选数
[CQOI2015]选数
https://ac.nowcoder.com/acm/problem/19940
[CQOI2015]选数
/* Author : lifehappy */ #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 2e6 + 10, mod = 1e9 + 7; int prime[N], mu[N], cnt; bool st[N]; ll quick_pow(ll a, int n) { ll ans = 1; while(n) { if(n & 1) ans = ans * a % mod; a = a * a % mod; n >>= 1; } return ans; } void init() { mu[1] = 1; for(int i = 2; i < N; i++) { if(!st[i]) { prime[++cnt] = i; mu[i] = -1; } for(int j = 1; j <= cnt && 1ll * 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] + mu[i - 1] + mod) % mod; } } unordered_map<int, int> ans_s; int Djs(int n) { if(n < N) return mu[n]; if(ans_s.count(n)) return ans_s[n]; int ans = 1; for(int l = 2, r; l <= n; l = r + 1) { r = n / (n / l); ans = (ans - 1ll * (r - l + 1) * Djs(n / l) % 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(); int N, K, L, H; scanf("%d %d %d %d", &N, &K, &L, &H); L = (L - 1) / K, H = H / K; //提前对 L 进行向下取整,所以在后面就可以不用进行ceil,然后相减加 1 操作了 ll ans = 0; for(int l = 1, r; l <= H; l = r + 1) { r = L / l ? min(L / (L / l), H / (H / l)) : H / (H / l); ans = (ans + (Djs(r) - Djs(l - 1)) * quick_pow(H / l - L / l, N) % mod + mod) % mod; } printf("%lld\n", ans); return 0; }