题面

Bzoj

COGS加强版

Sol

非加强版可以枚举AC这里不再讲述

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设\(f(i)\)表示在\([L, H]\)取\(N\)个，\(gcd为i\)的方案数

\(F(i)=\sum_{i|d}f(d)\)表示\([L,H]\)取\(N\)个，\(gcd为i\)的倍数的方案数

易得\(F(i)=(\lfloor\frac{H}{i}\rfloor-\lfloor\frac{L-1}{i}\rfloor)^N\)

直接莫比乌斯反演得到\(f(K)=\sum_{K|d}\mu(\frac{d}{K})F(d)\)

把\(\frac{d}{K}\)替换掉\(f(K)=\sum_{i=1}^{\lfloor\frac{H}{K}\rfloor}\mu(i)F(K*i)\)

分块\(F(K*i)\)杜教筛出\(\mu\)的前缀和就可以了

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(1e6 + 1), Zsy(1e9 + 7);

IL ll Read(){

    RG ll x = 0, z = 1; RG char c = getchar();

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);

    return x * z;

}

int prime[_], mu[_], s[_], num, L, H, N, K, MAXN;

map <int, int> Mu;

bool isprime[_];

IL ll Pow(RG ll x, RG ll y){

	RG ll ret = 1;

	for(; y; y >>= 1, x = x * x % Zsy) if(y & 1) ret = ret * x % Zsy;

	return ret;

}

IL void Prepare(){

	isprime[1] = 1; mu[1] = 1;

    for(RG int i = 2; i < MAXN; ++i){

        if(!isprime[i]){  prime[++num] = i; mu[i] = -1;  }

        for(RG int j = 1; j <= num && i * prime[j] < MAXN; ++j){

            isprime[i * prime[j]] = 1;

            if(i % prime[j]) mu[i * prime[j]] = -mu[i];

            else{  mu[i * prime[j]] = 0; break;  }

        }

		mu[i] += mu[i - 1];

    }

}

IL int SumMu(RG int n){

	if(n < MAXN) return mu[n];

	if(Mu[n]) return Mu[n];

	RG int ans = 1;

	for(RG int i = 2, j; i <= n; i = j + 1){

		j = n / (n / i);

		ans -= 1LL * (j - i + 1) * SumMu(n / i) % Zsy;

		ans = (ans + Zsy) % Zsy;

	}

	return Mu[n] = ans;

}

int main(RG int argc, RG char* argv[]){

	N = Read(); K = Read(); L = (Read() - 1) / K; H = Read() / K;

	RG int ans = 0, lst = 0, now; MAXN = min(H + 1, _); Prepare();

	for(RG int i = 1, j; i <= H; i = j + 1){

		j = H / (H / i); if(L / i) j = min(j, L / (L / i));

		now = SumMu(j);

		ans += 1LL * (now - lst) * Pow(H / i - L / i, N) % Zsy;

		ans = (ans % Zsy + Zsy) % Zsy; lst = now;

	}

	printf("%d\n", ans);

    return 0;

}