二分

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本文介绍了一种解决特殊二分查找问题的游戏算法，玩家通过选择序列中的两个数求和并取模获得分数，目标是找到能获得的最大分数。文章提供了一个C++实现示例，该算法首先对输入序列进行排序，然后采用二分查找的方法确定最优解。

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Pog and Szh are playing games.There is a sequence with n numbers, Pog will choose a number A from the sequence. Szh will choose an another number named B from the rest in the sequence. Then the score will be (A+B) mod p

.They hope to get the largest score.And what is the largest score?

Input

Several groups of data (no more than 5 groups,n≥1000).

For each case:

The following line contains two integers,n(2≤n≤100000)，p(1≤p≤231−1)。

The following line contains n integers ai(0≤ai≤231−1)

。

Output

For each case,output an integer means the largest score.

Sample Input

Sample Output

3
2

注意：特殊的二分

#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
    long long int n,p,x;
    long long int a[100000];
    int i,j;
    long long int r,l,mid;
    long long int max;
    while(scanf("%lld %lld",&n,&p)!=EOF)
    {
        for(i=0; i<n; i++)
        {
            scanf("%lld",&a[i]);
            //printf("%lld ",a[i]);
           if(a[i]>p)
                a[i]=a[i]%p;
        }
        sort(a,a+n);
        max=(a[n-1]+a[n-2])%p;
        l=0;
        r=n-1;
       // printf("\n%lld ",a[n-1]);
        while(r-l!=0)
        {
           // mid=(l+r)/2;
            x=(a[l]+a[r]);
            if(x%p>max)
                max=x%p;
                //printf("\n%lld",max);
            if(x>=p)
                r--;
            if(x<p)
                l++;
        }
        printf("%lld\n",max);

    }

    return 0;
}