div 794

​​https://codeforces.com/contest/1686​​

A

思路：数据量很小，100这样，直接暴力模拟O（n²）

/*
给你1--n，在你看了电影后，你站起来进行如下操作
操作1，你选择n-1个元素，替换掉他们成为他们的算数平均值
*/
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
const int maxn = 50 + 5;

int a[maxn] = {};

void s(){
  int n;
  cin >> n;
  
  for(int i = 1; i <= n; i++) cin >> a[i];
  
  sort(a+1, a+n+1);
  
  double sum = 0;
  for(int i = 1; i <= n ;i ++){
    sum = 0;
    for(int j = 1; j <= n ;j ++){
      if(j != i){
        sum += a[j];
      }      
    }
    sum /= (n - 1);
    if(sum == a[i]){
      cout << "YES\n";
      return ;
    }

  }
  
  cout << "NO\n";
}

int main(){

  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);

  #ifdef LOCAL
  FILE *p = fopen("input.txt", "a+");
  fclose(p);
  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
  #endif
  int t;
  cin >> t;
  while(t--) s();
  
  
  return 0;
}

B

思路：这题是要连续分割，不是挑出子区间。所以就找有几个断层点就行。比如345612，6与1之间就是断层。遇到断层就i+=2就行，继续往下分割

/*
一个序列，定义1--m中，bi》bj逆序，这样的逆序对总数是奇数的就叫奇
*/
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
const int maxn = 1e5 + 5;

void s(){
  int n ;
  cin >> n;
  int a[maxn] = {};
  memset(a, 0x3f, sizeof a);
  for(int i = 1; i <= n;i ++) cin >> a[i];
  
  long long ans =  0;
  for(int i = 1; i <= n; i ++){
    if(a[i] > a[i + 1]){
      ans ++;
      i++;
    }
  }
  
  cout << ans <<endl;
}

int main(){

  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);

  #ifdef LOCAL
  FILE *p = fopen("input.txt", "a+");
  fclose(p);
  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
  #endif
  int t;
  cin >>t;
  while(t--) s();
  
  
  return 0;
}

C

思路：稍微模拟一下。先排序，将整个数组从中切开，将前半部分分别放入13579..的位置，将后半部分放入2468...的位置，check是否满足奇数下标的小  偶数下标的大

/**/
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
const int maxn = 1e5 + 5;

int n;
int a[maxn];
int ans[maxn];

bool check(){
  
  if(ans[1] >= ans[n] || ans[n] <= ans[n - 1])
    return 0;
  
  for(int i = 2; i <= n - 2;i += 2){
    if(ans[i] <=ans[i - 1] || ans[i] <= ans[i + 1]){
      return 0;
    }
  }
  return 1;
}

void s(){
  cin >> n;
  
  for(int i = 1; i <= n; i ++) cin >> a[i];
  
  if(n % 2 == 0){
    sort(a+1, a+1+n);
    for(int i = 1, j = 1, k = n / 2 + 1; i <= n; i += 2){
      ans[i] = a[j++];
      ans[i + 1] = a[k++];
    }
    if(check()){
      cout << "YES\n";
      for(int i = 1; i <= n ;i ++) cout << ans[i] << " "; cout <<endl;
    }else cout << "NO\n";
  }else if(n % 2 == 1){
    cout << "NO\n";
  }
  


}

int main(){

  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);

  #ifdef LOCAL
  FILE *p = fopen("input.txt", "a+");
  fclose(p);
  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
  #endif
  int t; cin >> t;
  while(t--) s();
  
  
  return 0;
}