Codeforces Round ＃794 (Div. 2)(A.B.C)

A. Everything Everywhere All But One

Topic link ：

Problem - A - Codeforces

Topic ：

 

  The question ：

Yes n Number , You can choose n - 1 Numbers make these numbers here n-1 The arithmetic mean of the number . Ask if you can change these numbers into all the same values

Ideas ：

One n Number , You can choose n-1 Number , Then we can enumerate each number as the unselected number , Then see whether the arithmetic mean of other numbers is equal to the unselected value

Code ：

#include<bits/stdc++.h>
using namespace std;

int arr[55];

int main(){
	int t;
	cin >> t;
	while(t--){
		int n;
		cin >> n;
		double sum = 0;
		for(int i = 0; i < n; i++){
			cin >> arr[i];
			sum += arr[i];
		}
		bool f = 0; 
		for(int i = 0; i < n; i++){
			if((sum - arr[i]) / (n - 1) == arr[i]){
				f = 1;
				break;
			}
		}
		if(f){
			cout << "YES" << endl;
		}else{
			cout << "NO" << endl;
		}
	}
	return 0;
}

B. Odd Subarrays

Topic link ：

Problem - B - Codeforces

Topic ：

 

The question ：

An array b, If i<j also ai>aj, So this pair （i,j） Even a reversal number , If the number of inverses of this array is odd , Then say that this array is odd . Ask how many consecutive odd subarrays you can split an array into

Ideas ：

We ask for the most , that 2 The number of people in a group is the largest , If ai > ai+1, Then you can put these two together , Just divide the others into groups

Code ：

#include<bits/stdc++.h>
using namespace std;

int arr[100005];


int main(){
	int t;
	cin >> t;
	while(t--){
		int n;
		cin >> n;
		for(int i = 0; i < n; i++){
			cin >> arr[i];
		}
		int num = 0;
		for(int i = 0; i < n - 1; i ++){
			if(arr[i] > arr[i + 1]){
				num++;
				i++;
			}
		}
		cout << num << endl;
	}
	return 0;
}

C. Circular Local MiniMax

Topic link ：

Problem - C - Codeforces

Topic ：

The question ：

  Given some numbers , Change the position of the number , Ask whether these numbers can be enclosed into a garden , The intermediate value is required to be smaller than that of both adjacent walls , Or bigger than both sides

Ideas ：

If the number of these numbers is odd , Then you can never be satisfied , If it's even , Sort these Numbers from small to large , And then the front n/2 One and the back n/2 Alternate square . For the last time , Judge whether it will be the same

Code ：

#include<bits/stdc++.h>
using namespace std;

int arr[100005];
int b[100005];

int main(){
	int t;
	cin >> t;
	while(t--){
		int n;
		cin >> n;
		for(int i = 0; i < n; i++){
			cin >> arr[i];
		}
		sort(arr, arr + n);
		if(n & 1){
			cout << "NO" << endl;
			continue;
		}else{
			int a = 0;
			for(int i = 0; i < n / 2; i++){
				b[a] = arr[i];
				a++;
				b[a] = arr[n / 2 + i];
				a++;
			}
		}
		bool f = 0;
		for(int i = 1; i < n - 1; i++){
			if(b[i] == b[i - 1] || b[i] == b[i + 1]){
				f = 1;
			}
		}
		if(b[0] == b[n - 1]){
			f = 1;
		}
		if(f){
			cout << "NO" << endl;
		}else{
			cout << "YES" << endl;
			for(int i = 0; i < n; i++){
				if(i != 0){
					cout << " ";
				}
				cout << b[i];
			}
			cout << endl;
		}
	}
	return 0;
}