本文共 1638 字，大约阅读时间需要 5 分钟。

一.题目链接：

二.题目大意：

中文题~~

三.分析：

正如题解所述，一步一步分析即可.

计算和式时，分情况讨论即可.

四.代码实现：

#include
   
    using namespace std;typedef long long ll;const int M = (int)1e5;const ll mod = (ll)1e9 + 7;int l[M + 5];int r[M + 5];ll quick(ll a, int b){    ll sum = 1;    while(b)    {        if(b & 1)   sum = sum * a % mod;        a = a * a % mod;        b >>= 1;    }    return sum;}ll inv(ll n){    return quick(n, mod - 2);}int main(){    int n; scanf("%d", &n);    ll mul = 1; for(int i = 1; i <= n; ++i) scanf("%d %d", &l[i], &r[i]), mul = mul * (r[i] - l[i] + 1) % mod;    ll ans = mul;    for(int i = 2; i <= n; ++i)    {        ll sum = 0;        if(l[i - 1] < l[i] && l[i] < r[i - 1] && r[i - 1] < r[i])       sum = 1ll * (r[i - 1] - l[i]) * (r[i - 1] - l[i] + 1) / 2 % mod;        if(l[i - 1] <= l[i] && l[i] <= r[i] && r[i] <= r[i - 1])        sum = 1ll * (r[i] - l[i] + 1) * (2 * r[i - 1] - l[i] - r[i]) / 2 % mod;        if(l[i] < l[i - 1] && l[i - 1] <= r[i] && r[i] < r[i - 1])      sum = (1ll * (l[i - 1] - l[i]) * (r[i - 1] - l[i - 1] + 1) % mod + 1ll * (r[i] - l[i - 1] + 1) * (2 * r[i - 1] - l[i - 1] - r[i]) / 2 % mod) % mod;        if(l[i] <= r[i] && r[i] < l[i - 1] && l[i - 1] <= r[i - 1])     sum = 1ll * (r[i - 1] - l[i - 1] + 1) * (r[i] - l[i] + 1) % mod;        if(l[i] < l[i - 1] && l[i - 1] <= r[i - 1] && r[i - 1] < r[i])  sum = (1ll * (l[i - 1] - l[i]) * (r[i - 1] - l[i - 1] + 1) % mod + 1ll * (r[i - 1] - l[i - 1]) * (r[i - 1] - l[i - 1] + 1) / 2 % mod) % mod;        sum = sum * mul % mod * inv(r[i] - l[i] + 1) % mod * inv(r[i - 1] - l[i - 1] + 1) % mod;        ans = (ans + sum) % mod;    }    printf("%lld\n", ans);    return 0;}
   

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