思路：
https://codeforces.com/blog/entry/68111

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
     
      using namespace std;#define y1 y11#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
      
       #define pli pair
       
        #define pii pair
        
         #define piii pair
         
          #define pdd pair
          
           #define mem(a, b) memset(a, b, sizeof(a))#define debug(x) cerr << #x << " = " << x << "\n";#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//headconst int MOD = 1e9 + 7;const int N = 2e5 + 5;int l[N], r[N], n;LL p[N], q[N], sum[N];LL q_pow(LL n, LL k) { LL res = 1; while(k) { if(k&1) res = (res * n) % MOD; n = (n * n) % MOD; k >>= 1; } return res;}int main() { scanf("%d", &n); for (int i = 1; i <= n; ++i) scanf("%d", &l[i]); for (int i = 1; i <= n; ++i) scanf("%d", &r[i]); for (int i = 1; i <= n; ++i) { int L = max(l[i-1], l[i]), R = min(r[i-1], r[i]); if(L > R) p[i] = 1, q[i] = 0; else q[i] = (R-L+1)*q_pow((r[i]-l[i]+1)*1LL*(r[i-1]-l[i-1]+1)%MOD, MOD-2)%MOD, p[i] = (1-q[i]+MOD)%MOD; sum[i] = (sum[i-1] + p[i]) % MOD; } LL ans = sum[n]; for (int i = 1; i <= n ;++i) { LL tot = sum[n]; tot = (tot - p[i])%MOD; if(i-1 >= 1) tot = (tot - p[i-1]) % MOD; if(i+1 <= n) tot = (tot - p[i+1]) % MOD; tot = (tot + MOD) % MOD; ans = (ans + p[i]*tot%MOD) % MOD; } for (int i = 1; i < n; ++i) { LL tot = ((1-q[i]-q[i+1])%MOD+MOD)%MOD; if(i-1 >= 1) { int L = max(l[i-1], max(l[i], l[i+1])), R= min(r[i-1], min(r[i], r[i+1])); if(L <= R)tot = (tot + (R-L+1)*q_pow((r[i]-l[i]+1)*1LL*(r[i-1]-l[i-1]+1)%MOD*(r[i+1]-l[i+1]+1)%MOD, MOD-2)%MOD)%MOD; } ans = (ans + 2*tot) % MOD; } printf("%lld\n", ans%MOD); return 0;}