美丽的序列I（牛客小白月赛23 F，计数）

最新推荐文章于 2023-09-24 17:06:39 发布

The___Flash

最新推荐文章于 2023-09-24 17:06:39 发布

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本文链接：https://blog.youkuaiyun.com/The___Flash/article/details/105046008

本文详细解析了一道算法题目“美丽的序列I”的解题思路与代码实现过程，通过分情况讨论计算和式，运用快速幂运算提高效率，并展示了完整的C++代码实现。

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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;
}