Problem Description

You are given an integer sequence of length N, a1,a2,…,aN.

For each 1≤i≤N, you have three choices: add 1 to ai, subtract 1 from ai or do nothing.

After these operations, you select an integer X and count the number of i such that ai=X.

Maximize this count by making optimal choices.

Constraints

• 1≤N≤105
• 0≤ai<105(1≤i≤N)
• ai?is an integer.

Input

The input is given from Standard Input in the following format:

N
a1 a2 .. aN

Output

Print the maximum possible number of i such that ai=X.

Example

Sample Input 1

7
3 1 4 1 5 9 2

Sample Output 1

4
For example, turn the sequence into 2,2,3,2,6,9,2 and select X=2 to obtain 4, the maximum possible count.

Sample Input 2

10
0 1 2 3 4 5 6 7 8 9

Sample Output 2

3

Sample Input 3

1
99999

Sample Output 3

1

題意：給一個長度為 n 的整數序列，對于序列中的每一個數可以進行一次操作，可以對其 +1、-1 或不做任何操作，對這 n 個數進行一次操作后，問序列中次數出現最多的數

思路：由于給出的數 ai 的值不大，可以利用桶排，對于每個數將其 +1、-1、原值均存入桶中，最后掃一遍桶，桶中元素最大值就是答案，需要注意的是，由于 ai 最小可能為 0，將其 -1 后值為 -1，因此桶需要整體向右偏移一個

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 100000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;int bucket[N]; int main() {int n;scanf("%d",&n);for(int i=1;i<=n;i++){int x;scanf("%d",&x);x++;//向右偏移bucket[x]++;bucket[x-1]++;bucket[x+1]++;}int maxx=-INF;int res=0;for(int i=1;i<=1E5+1;i++)maxx=max(maxx,bucket[i]);printf("%d\n",maxx);return 0; }

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