Codeforces 1187F Expected Square Beauty （看题解）

Expected Square Beauty

感觉是个处理平方期望的套路题。。

看题解就好啦。

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

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-10;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());


int power(int a, int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL * ans * a % mod;
        a = 1LL * a * a % mod; b >>= 1;
    }
    return ans;
}

int n, l[N], r[N];

int p[N], q[N];
int pre[N];

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]);
        r[i]++;
    }

    q[1] = 0; p[1] = 1;

    for(int i = 1; i <= n; i++) {

        int L = max(l[i], l[i - 1]);
        int R = min(r[i], r[i - 1]);

        if(R > L) {

            q[i] = 1LL * (R - L) * power(r[i] - l[i], mod - 2) % mod * power(r[i - 1] - l[i - 1], mod - 2) % mod;
        }

        p[i] = (1 - q[i] + mod) % mod;
    }

    for(int i = 1; i <= n; i++) {

        pre[i] = (pre[i - 1] + p[i]) % mod;
    }

    int ans = 0;

    // i == j
    for(int i = 1; i <= n; i++) {

        add(ans, p[i]);
    }


    // abs(j - i) > 1
    for(int i = 1; i <= n; i++) {

        if(i - 2 >= 1) {

            add(ans, 2LL * p[i] * pre[i - 2] % mod);
        }
    }

    r[0] = -inf; l[0] = inf;

    // abs(j - i) == 1
    for(int i = 1; i < n; i++) {

        int ret = ((1 - q[i] - q[i + 1]) % mod + mod) % mod;

        int R = min(r[i - 1], min(r[i], r[i + 1]));
        int L = max(l[i - 1], max(l[i], l[i + 1]));

        int tmp = 0;

        if(R > L) {

            tmp = R - L;
            tmp = 1LL * tmp * power(r[i - 1] - l[i - 1], mod - 2) % mod;
            tmp = 1LL * tmp * power(r[i] - l[i], mod - 2) % mod;
            tmp = 1LL * tmp * power(r[i + 1] - l[i + 1], mod - 2) % mod;
        }

        add(ret, tmp);
        add(ans, 2LL * ret % mod);


    }

    printf("%d\n", ans);

    return 0;
}

/*
*/