CodeForces 653 A. Bear and Three Balls（数学 ，快排）

Description
Limak is a little polar bear. He has n balls, the i-th ball has size ti.

Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:

No two friends can get balls of the same size.
No two friends can get balls of sizes that differ by more than 2.
For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).

Your task is to check whether Limak can choose three balls that satisfy conditions above.

Input
The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.

The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.

Output
Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).

Sample Input
Input
4
18 55 16 17
Output
YES
Input
6
40 41 43 44 44 44
Output
NO
Input
8
5 972 3 4 1 4 970 971
Output

YES

代码：

#include<iostream>#include<stdio.h>#include<math.h>#include<algorithm>#include<string.h>using namespace std;int main(){int n;while(scanf("%d",&n)!=EOF){int a[51],k=0;memset(a,0,sizeof(a));for(int i=1;i<=n;i++){scanf("%d",&a[i]);if(a[i]==a[i-1]){a[i-1]=-1;}}sort(a+1,a+n+1);for(int i=1;i<=n;i++){if(a[i]==a[i-1]){a[i-1]=-1;k++;}}sort(a+1,a+n+1);int ans=0;for(int i=k+1;i<=n-2;i++){if(a[i+2]-a[i+1]==1&&a[i+1]-a[i]==1){ans++;printf("YES\n");break;}}if(ans==0)printf("NO\n"); } return 0;}

题意：小熊要送给三个小伙伴篮球。要求任意两个球不相同而且价值差不大于2；给你篮球的我价值，问小熊是否能送成功。

思路：就是判断有没有三个球连续。输入篮球价值后快排，这时要将相同价值的篮球消去，不然后来判断是否连续比较麻烦。然后直接判断三个球是否连续就行了

a[i+2]-a[i+1]==1&&a[i+1]-a[i]==1。（这道题wa了我三次，开始是没有考虑相同的值，然后是忘记输出YES后要break结果就是有的数据输出一堆YES，第三次还是消除相同的值没有做好晕啊）