[CQOI2015]选数

[CQOI2015]选数

https://ac.nowcoder.com/acm/problem/19940

[CQOI2015]选数

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/*
  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;
}
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