AtCoder Beginer Contest 236 E

E.Average and Median 题目大意

给定 N N N张卡牌，每张卡牌上有一个权值 A i A_i Ai​，现在要求从这 N N N张卡片种选择一些卡片，要求对于 ∀ i ∈ [ 1 , N ] \forall i \in [1, N] ∀i∈[1,N]，必须有第 i i i张或 i + 1 i+1 i+1张卡牌被选择。求在所有选择方案中：①.所选卡牌的最大平均值； ②.所选卡牌的最大中位数

思路

毫无思路直接在值域上二分答案然后检验即可。

竟然没想到

Accepted Code

#include 
#define endl '\n'
using namespace std;

const int N = 1e5 + 10;
const double det = 1e-5;
int n，a[N], b[N];
double dp[N];

inline bool CheckAve(double mid){
    for(int i = 1; i = 0;
}

inline bool CheckMid(int mid){
    for(int i = 1; i = mid) ? 1 : -1;
    for(int i = 1; i  0;
}

inline void solve(){
    cin >> n;
    for(int i = 1; i > a[i];
    double l1 = 0, r1 = 1e9, ans1 = 0;
    while(r1 - l1 > det){
        double mid = (l1 + r1) / 2;
        if(CheckAve(mid)) l1 = ans1 = mid;
        else r1 = mid;
    }
    int l = 1, r = 1e9, ans2 = 0;
    while(l > 1;
        if(CheckMid(mid)) ans2 = mid, l = mid + 1;
        else r = mid - 1;
    }
    cout