题目链接
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 = sumlimits_{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
",ans);
return 0;
}