一、Remove Smallest

本题链接：Remove Smallest

题目：
A. Remove Smallest
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given the array a consisting of n positive (greater than zero) integers.

In one move, you can choose two indices i and j (i≠j) such that the absolute difference between ai and aj is no more than one (|ai−aj|≤1) and remove the smallest of these two elements. If two elements are equal, you can remove any of them (but exactly one).

Your task is to find if it is possible to obtain the array consisting of only one element using several (possibly, zero) such moves or not.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤1000) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1≤n≤50) — the length of a. The second line of the test case contains n integers a1,a2,…,an (1≤ai≤100), where ai is the i-th element of a.

Output
For each test case, print the answer: “YES” if it is possible to obtain the array consisting of only one element using several (possibly, zero) moves described in the problem statement, or “NO” otherwise.

Example
input
5
3
1 2 2
4
5 5 5 5
3
1 2 4
4
1 3 4 4
1
100
output
YES
YES
NO
NO
YES
Note
In the first test case of the example, we can perform the following sequence of moves:

choose i=1 and j=3 and remove ai (so a becomes [2;2]);
choose i=1 and j=2 and remove aj (so a becomes [2]).
In the second test case of the example, we can choose any possible i and j any move and it doesn’t matter which element we remove.

In the third test case of the example, there is no way to get rid of 2 and 4.

本博客给出本题截图：

题意：给t组数字串，对于每组数字有两种操作：挑选两个绝对值相差1的数，并删除其中的小数；挑选两个值相同的数字并任意删除一个数

AC代码

#include <iostream>
#include <algorithm>using namespace std;const int N = 60;int a[N];int main()
{int t;cin >> t;while (t -- ){int n;cin >> n;for (int i = 0; i < n; i ++ ) cin >> a[i];if (n == 1){puts("YES");continue;}sort(a, a + n);int cnt = 0;for (int i = 0; i < n - 1; i ++ ) if (a[i + 1] - a[i] == 1 || a[i + 1] == a[i])cnt ++;if (cnt == n - 1) puts("YES");else puts("NO");}return 0;
}

总结

水题，不解释

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Remove Smallest

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一、Remove Smallest

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