洛谷 P3172 ：[CQOI2015]选数（莫比乌斯反演 + 杜教筛）

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$$\sum _{i_1=1}^h\sum _{i_2=1}^h\sum _{i_3=1}^h...[gcd(i_1,i_2,i_3,..)=k]$$$$=\sum _{i_1=1}^{\lfloor \frac{h}{k}\rfloor }\sum _{i_2=1}^{\lfloor \frac{h}{k}\rfloor }\sum _{i_3=1}^{\lfloor \frac{h}{k}\rfloor }...[gcd(i_1,i_2,i_3,..)=1]$$$$=\sum _{p=1}^{\lfloor \frac{h}{k}\rfloor }\mu (p)\sum _{i_1=1}^{\lfloor \frac{h}{k\ast p}\rfloor }\sum _{i_2=1}^{\lfloor \frac{h}{k\ast p}\rfloor }\sum _{i_3=1}^{\lfloor \frac{h}{k\ast p}\rfloor }...$$$$=\sum _{p=1}^{\lfloor \frac{h}{k}\rfloor }\mu (p)\ast (\lfloor \frac{h}{k\ast p}\rfloor -\lfloor \frac{l-1}{k\ast p}\rfloor {)}^n$$

右边可以分块，需要预处理 $\mu (i)$的前缀和，由于 上限达到了 $10^9$，上杜教筛

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#include
using namespace std;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
bool ispri[maxn];
int pri[maxn],mu[maxn];
int sum[maxn];
map<int,int> mp;
int n,k,l,h;
inline int read(){
    int x=0,f=1; char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar())  if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar())    x=(x<<3)+(x<<1)+ch-'0';
    return x*f;
}
void sieve(int n) {
	ispri[0] = ispri[1] = true;
	pri[0] = 0;mu[1] = 1;
	for(int i = 2; i <= n; i++) {
		if(!ispri[i]) pri[++pri[0]] = i,mu[i] = -1;
		for(int j = 1; j <= pri[0] && i * pri[j] <= n; j++) {
			ispri[i * pri[j]] = true;
			if(i % pri[j] == 0) break;
			mu[i * pri[j]] = -mu[i];
		}
	}
	sum[0] = 0;
	for(int i = 1; i <= n; i++)
		sum[i] = (sum[i - 1] + mu[i]) % mod;
}
int getsum(int x) {
	if(x <= maxn - 10) return sum[x];
	if(mp[x]) return mp[x];
	int res = 1;
	int i,j;
	for(i = 2; i <= x; i = j + 1) {
		j = x / (x / i);
		res = (res - 1ll * (j - i + 1) * getsum(x / i) % mod + mod) % mod;
	}
	return mp[x] = res;
}
int fpow(int a,int b) {
	int r = 1;
	while(b) {
		if(b & 1) r = 1ll * r * a % mod;
		a = 1ll * a * a % mod;
		b >>= 1;
	}
	return r;
}
int main() {
	sieve(maxn - 10);	
	n = read();k = read();l = read();h = read();
	h = h / k;
	l = (l - 1) / k;
	int i,j;
	int ans = 0;
	for(i = 1; i <= h; i = j + 1) {
		if(i <= l) j = min(h / (h / i),l / (l / i));
		else j = h / (h / i);
		int tmp = fpow((h / i - l / i + mod) % mod,n);
		ans = (ans + 1ll * ((getsum(j) - getsum(i - 1) + mod) % mod) * tmp % mod) % mod;	
	}
	printf("%d\n",ans);
	return 0;
}