[cf] Codeforces Round #794 (Div. 2) A-C

前言

熬夜掉分

A.

题意 : 给定你一个数组,询问是否可以通过有限次操作,使得数组的值相同

操作定义如下 : 选择数组中 ( n − 1 ) (n-1) (n−1)个数,将选中的每个数变为这些数的平均数(不进行取整操作)

思路:

证明不来

code :

int a[N];
int n;

void solve(){
          
   
	cin>>n;
	int sum = 0 ;
	
	for(int i=1;i<=n;i++) cin>>a[i],sum+=a[i];
	
	for(int i=1;i<=n;i++){
          
   
		int t = (sum - a[i])/(n-1);
		if(t*(n-1) != sum - a[i] ) continue;
		
		if(t == a[i]){
          
   
			//cout<<t<<" "<<a[i]<<endl;
			
			cout<<"YES"<<endl;
			return;
		}
	}
	cout<<"NO"<<endl;
	
}

B.

题意 :

思路 :

code :

int n;
int a[N];

void solve(){
          
   
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	
	int res = 0 ;
	for(int i=2;i<=n;){
          
   
		if(a[i] < a[i-1]){
          
   
			++res;
			i+=2;
		}else i ++ ;
	} 
	
	cout<<res<<endl;
	
}

C.

题意 :

思路 :

code :

int n;
int a[N];
int ans[N];

void solve(){
          
   
	cin>>n;
	for(int i = 1;i<=n;i++ ) cin>>a[i];
	sort(a+1,a+1+n);
	
	int k = 0;
	
	
	
	for(int i = 1,j = n/2+1 ; i<=n/2; i ++ , j ++ ){
          
   
		ans[++k] = a[i];
		ans[++k] = a[j];
	}
	
	
	if(n&1){
          
   
		cout<<"NO"<<endl;
		return;
	}
	
	
	//b[0] = b[k]
	//b[k+1] = b[1]
	
	ans[0] = ans[k];
	ans[k+1] = ans[1];
	for(int i = 1 ; i <= k ; i++ ){
          
   
		if(ans[i] > ans[i-1] && ans[i] > ans[i+1]) continue;
		if(ans[i] < ans[i-1] && ans[i] < ans[i+1]) continue;
		cout<<"NO"<<endl;
		return;
	}
	
	cout<<"YES"<<endl;
	for(int i=1;i<=n;i++)
	cout<<ans[i]<<" ";
	cout<<endl;
	
	
	
}