题意

题目链接

Sol

这个题就比较休闲了。

\(t(p)\)显然等于最后一个没有约数的数的位置，那么我们可以去枚举一下。

设没有约数的数的个数有\(cnt\)个

因此总的方案为\(\sum_{i=cnt}^{r-l+1} C_{i-1}^{cnt-1} cnt! (r - l + 1 - cnt)!\)

稍微有点卡常，筛的时候加一下剪枝

#include<bits/stdc++.h>

#define Fin(x) freopen(#x".in", "r", stdin);

using namespace std;

const int MAXN = 1e7 + 10, mod = 1e9 + 7;

template<typename A, typename B> inline bool chmax(A &x, B y) {return x < y ? x = y, 1 : 0;}

template<typename A, typename B> inline bool chmin(A &x, B y) {return x > y ? x = y, 1 : 0;}

template<typename A, typename B> inline A mul(A x, B y) {return 1ll * x * y % mod;}

template<typename A, typename B> inline void add2(A &x, B y) {x = x + y >= mod ? x + y - mod : x + y;}

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

	return x * f;

}

int fac[MAXN], ifac[MAXN], vis[MAXN], cnt;

int fp(int a, int p) {

	int base = 1;

	while(p) {

		if(p & 1) base = mul(base, a);

		a = mul(a, a);  p >>= 1;

	}

	return base;

}

int C(int N, int M) {

	return mul(fac[N], mul(ifac[M], ifac[N - M]));

}

int main() {

	int l = read(), r = read();

	for(int i = l; i <= r; i++) {

		if(vis[i]) continue;

		if(!vis[i]) cnt++;

		for(int j = i + i; j <= r; j += i)

			vis[j] = 1;

	}

	fac[0] = 1;

	for(int i = 1; i <= r; i++) fac[i] = mul(i, fac[i - 1]);

	ifac[r] = fp(fac[r], mod - 2);

	for(int i = r; i; i--) ifac[i - 1] = mul(ifac[i], i);

	int ans = 0;

	for(int i = cnt; i <= r - l + 1; i++)

		add2(ans, mul(i, mul(C(i - 1, cnt - 1), mul(fac[cnt], fac[r - l + 1 - cnt]))));

	cout << ans;

	return 0;

}