二分

ABC 236 E - Average and Median

题目链接
https://atcoder.jp/contests/abc236/tasks/abc236_e

解析

• 想到了dp以及状态表示，但是没想出来怎么状态转移，主要没想到是用dp来判断mid行不行的，总体框架来看是个二分答案的题
• 本题对于中位数和平均数的处理特别经典，可以学习
• 题解就不写了，参考这篇博客吧
  https://blog.csdn.net/Mr_dimple/article/details/122679274

Ac代码

点击查看代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
const int N = 1e5 + 10;
 
int n;
double f[N][2];
double b[N];
int a[N];
 
bool check1(double mid){
    for(int i = 1; i <= n; i ++) b[i] = a[i] - mid;
 
    for(int i = 1; i <= n; i ++){
        f[i][0] = f[i - 1][1];
        f[i][1] = max(f[i - 1][0], f[i - 1][1]) + b[i];
    }
    return max(f[n][0], f[n][1]) > 0;
}
 
bool check2(int mid){
    for(int i = 1; i <= n; i ++) b[i] = a[i] >= mid ? 1 : -1;
 
    for(int i = 1; i <= n; i ++){
        f[i][0] = f[i - 1][1];
        f[i][1] = max(f[i - 1][1], f[i - 1][0]) + b[i];
    }
    return max(f[n][0], f[n][1]) > 0;
}
 
int main()
{
    scanf("%d", &n);
    int maxa = 0;
    for(int i = 1; i <= n; i ++) scanf("%d", &a[i]), maxa = max(maxa, a[i]);
 
    //求最大平均数
    double ll = 0, rr = maxa;
    while(rr - ll > 1e-4){
        double mid = (ll + rr) / 2;
        if(check1(mid)) ll = mid;
        else rr = mid;
    }
    printf("%.10lf\n", ll);
 
    //求最大中位数
    int l = 0, r = 1e9;
    while (l < r)
    {
        int mid = l + r + 1 >> 1;
        if (check2(mid)) l = mid;
        else r = mid - 1;
    }
    printf("%d\n", l);
 
    return 0;
}