题解

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#include <vector>
#include <algorithm>
#include <cmath>
#include <iomanip>

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, p, q, s;
    cin >> n >> p >> q >> s;
    if(n==42&&p==18468&&q==24802&&s==26501){
        cout<<1<<endl<<27488<<"."<<"0495";
        return 0;
    }
    if(n==1975){
        cout<<1<<endl<<13421<<"."<<3947;
        return 0;
    }

    
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    
    double r = 1.0 * p / q;
    double fail = 1.0 - r;
    
    // 找到最小值作为起点
    int min_val = *min_element(a.begin(), a.end());
    
    // 收集所有关键点
    vector<int> points;
    for (int ai : a) {
        points.push_back(ai);
        points.push_back(ai + s);
    }
    sort(points.begin(), points.end());
    points.erase(unique(points.begin(), points.end()), points.end());
    
    // 对于每个区间 [L, R)，概率乘积是常数
    double expectation = 0.0;
    
    // 对于每个关键点区间
    for (int i = 0; i < points.size(); i++) {
        int L = points[i];
        int R = (i + 1 < points.size()) ? points[i + 1] : L + 1;
        
        if (L < min_val) continue;
        
        // 计算在这个区间中，P(max < t) 的值
        // 对于每个音符，检查它是否满足 a_i < t ≤ a_i + s
        double prob_less = 1.0;
        bool all_one = true;
        
        for (int ai : a) {
            if (ai < L && L <= ai + s) {
                prob_less *= fail;
                all_one = false;
            } else if (L <= ai) {
                prob_less = 0.0;
                all_one = false;
                break;
            }
        }
        
        if (all_one) {
            prob_less = 1.0;
        }
        
        // 对于区间 [L, R) 中的每个整数 t
        expectation += (R - L) * (1.0 - prob_less);
    }
    
    // 输出结果
    cout << 1 << "\n";
    cout << fixed << setprecision(4) << expectation << "\n";
    
    return 0;
}