一.题目链接:
美丽的序列I
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<bits/stdc++.h>
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;
}