B. Wall painting (C language)

subject

Ideas

First , From the title we can get ：
······ Two numbers with the same parity can have the same value by adding building blocks
······ Two adjacent equal numbers can change parity at the same time （ That is to add a horizontal building block to it ）
thus , We can get ：
······ Odd numbers two numbers with the same parity cannot become exactly equal numbers
······ Even numbers two numbers with the same parity can become exactly equal numbers
therefore ：
······ When the number between two numbers with the same parity is even , Then these two numbers and each of them become equal numbers
······ When the number between two numbers with the same parity is an odd number , Then these two numbers and each of them cannot be completely equal
So the problem becomes ：
······ Judge whether there are even numbers between two numbers with the same parity , If it is , Then the walls in this area can be the same height （ In other words, the wall in this area can be any height , Adding blocks can change the height ）, On the contrary, we can't .
Concrete realization ：
There are two cases ：
Finally, the height of the wall becomes an odd number ： Find two nodes with odd numbers , Judge whether the number between them is even , If it is , Continue to look for the following two odd nodes , Continue to judge , Until all are even numbers , Walls can be the same height ; If not , The walls can't be the same height .
Even number homology .
In the case of even numbers and odd numbers, only one kind of , The wall can be the same height
Be careful ： Someone might ask , How to calculate the odd node between the current two odd nodes and the following two odd nodes , Actually , Add several building blocks directly vertically , Can be any odd number ; Even nodes are the same . So these nodes don't matter .

Code

#include<stdio.h>

int n;
int a[500000];

int judge(int x);
int main()
{
	
	
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	if(judge(1)||judge(0)){
		printf("YES");
	}else{
		printf("NO");
	}
	
 } 
 
int judge(int x)
{
	int index=-1;
	for(int i=0;i<n;i++){
		if(a[i]%2==x){
			continue;
		}
		if(index!=-1){
			int len=i-index-1;
			if(len%2==1){
				return 0;
			}
			index=-1;
		}else{
			index=i;
		}
	}
	return index==-1;
}