A. DZY Loves Hash

最新推荐文章于 2019-03-06 20:57:37 发布

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最新推荐文章于 2019-03-06 20:57:37 发布

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分类专栏： Codeforces_Div.2专题文章标签： CF_Div. 2_#254

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本文介绍了一个简单的哈希表应用案例，通过插入一系列数值并检测哈希冲突来确定第一个发生冲突的位置。使用C++实现了一个基本的哈希表，并演示了如何通过取模运算将数值映射到哈希表中。

DZY has a hash table with p buckets, numbered from0 to p - 1. He wants to insertn numbers, in the order they are given, into the hash table. For thei-th number x_i, DZY will put it into the bucket numberedh(x_i), whereh(x) is the hash function. In this problem we will assume, thath(x) = x mod p. Operationa mod b denotes taking a remainder after divisiona by b.

However, each bucket can contain no more than one element. If DZY wants to insert an number into a bucket which is already filled, we say a "conflict" happens. Suppose the first conflict happens right after thei-th insertion, you should output i. If no conflict happens, just output -1.

Input

The first line contains two integers, p andn (2 ≤ p, n ≤ 300). Thenn lines follow. The i-th of them contains an integer x_i(0 ≤ x_i ≤ 10⁹).

Output

Output a single integer — the answer to the problem.

Sample test(s)

Input

Output

Input

Output

-1

//水题不多说

//AC代码

#include<iostream>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
using namespace std;
int main()
{
    int p,n,x,y,j;
    int z[301];
    //while(1)
    //{
        cin>>p>>n;
        map<int,int>Map;
        memset(z,0,sizeof(z));
        j=0;
        for(int i=1;i<=n;i++)
        {
            cin>>x;
            y=x%p;
            Map[y]+=1;
            if(Map[y]>1)
            {
                z[j++]=i;
            }
        }
        if(j!=0)
        {
            cout<<z[0]<<endl;
        }
        else
            cout<<-1<<endl;
    //}
    return 0;
}