Together（AtCoder-3524）

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 個數進行一次操作後，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，對于每個數将其 +1、-1、原值均存入桶中，最後掃一遍桶，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，将其 -1 後值為 -1，是以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;
}