BZOJ5323 [Jxoi2018]游戏 【数论/数学】

题目链接

BZOJ5323

题解

有一些数是不能被别的数筛掉的

这些数出现最晚的位置就是该排列的\(t(p)\)

所以我们只需找出所有这些数，线性筛一下即可，设有\(m\)个

然后枚举最后的位置

\[ans = \sum\limits_{i = m}^{n} m!(n - m)!{i - 1 \choose m - 1}i
\]

复杂度\(O(n)\)

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 10000005,P = 1000000007;
int p[maxn],isn[maxn],pi,L,R,len;
int fac[maxn],fv[maxn],inv[maxn];
void pre(){
	fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1;
	for (register int i = 2; i <= len; i++){
		fac[i] = 1ll * fac[i - 1] * i % P;
		inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
		fv[i] = 1ll * fv[i - 1] * inv[i] % P;
	}
}
void init(){
	for (register int i = 2; i <= R; i++){
		if (!isn[i]) p[++pi] = i;
		for (register int j = 1; j <= pi && i * p[j] <= R; j++){
			isn[i * p[j]] = p[j];
			if (i % p[j] == 0) break;
		}

	}
}
int main(){
	scanf("%d%d",&L,&R); len = R - L + 1;
	pre();
	int cnt = 0;
	if (L == 1) cnt = 1;
	else{
		init();
		for (register int i = L; i <= R; i++){
			if (!isn[i] || i / isn[i] < L)
				cnt++;
		}
	}
	int ans = 0;
	for (register int i = cnt; i <= len; i++){
		ans = (ans + 1ll * fac[i - 1] % P * fv[i - cnt] % P * i % P) % P;
	}
	ans = 1ll * ans * cnt % P * fac[len - cnt] % P;
	printf("%d\n",ans);
	return 0;
}