微商货源网站大全,百度引擎的搜索方式是什么,网站开发 h5,网站让百度收录应该怎么做题意#xff1a;
现有一个定义#xff1a;#xff08;1#xff09;两个数的奇偶性相同 #xff08;2#xff09;两个数的差的绝对值为1 #xff0c;即|a-b|1 满足以上两个条件之一#xff0c;就可以说两个数是相似的 先给你一个数组#xff0c;让你将数组内分成多组
现有一个定义1两个数的奇偶性相同 2两个数的差的绝对值为1 即|a-b|1 满足以上两个条件之一就可以说两个数是相似的 先给你一个数组让你将数组内分成多组每组两个元素是否都相似。
题目
We call two numbers x and y similar if they have the same parity (the same remainder when divided by 2), or if |x−y|1. For example, in each of the pairs (2,6), (4,3), (11,7), the numbers are similar to each other, and in the pairs (1,4), (3,12), they are not.
You are given an array a of n (n is even) positive integers. Check if there is such a partition of the array into pairs that each element of the array belongs to exactly one pair and the numbers in each pair are similar to each other.
For example, for the array a[11,14,16,12], there is a partition into pairs (11,12) and (14,16). The numbers in the first pair are similar because they differ by one, and in the second pair because they are both even.
Input The first line contains a single integer t (1≤t≤1000) — the number of test cases. Then t test cases follow.
Each test case consists of two lines.
The first line contains an even positive integer n (2≤n≤50) — length of array a.
The second line contains n positive integers a1,a2,…,an (1≤ai≤100).
Output
For each test case print:
YES if the such a partition exists, NO otherwise. The letters in the words YES and NO can be displayed in any case.
Example
Input
7 4 11 14 16 12 2 1 8 4 1 1 1 1 4 1 2 5 6 2 12 13 6 1 6 3 10 5 8 6 1 12 3 10 5 8
Output
YES NO YES YES YES YES NO
Note
The first test case was explained in the statement.
In the second test case, the two given numbers are not similar.
In the third test case, any partition is suitable.
分析
由于n为偶数那么奇数个数和偶数个数奇偶性相同 1如果个数都为偶直接各自配对 2如果个数为奇我们只需要判断是否存在一对奇数和偶数相差为1配对如果找不到就NO
AC代码
#includestdio.h
#includecstring
#includealgorithm
using namespace std;
int t,n,x,y,flag;
int dp[60];
int main()
{scanf(%d,t);while(t--){xyflag0;scanf(%d,n);for(int i0; in; i){scanf(%d,dp[i]);if(dp[i]%2) x;}if(x%20)flag1;sort(dp,dpn);for(int i1; in; i)if(dp[i]-dp[i-1]1){flag1;break;}flag1?printf(YES\n):printf(NO\n);}return 0;
}
