Codeforces Round #FF (Div. 2) A. DZY Loves Hash

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本文介绍了一个简单的哈希表冲突检测算法。通过将一系列整数插入到具有固定桶数量的哈希表中，并使用模运算作为哈希函数，来确定首次出现冲突的位置。如果在尝试插入相同哈希值的元素时发生冲突，则输出该冲突发生在第几个插入操作。

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DZY has a hash table with p buckets, numbered from 0 to p - 1. He wants to insert n numbers, in the order they are given, into the hash table. For the i-th number xi, DZY will put it into the bucket numbered h(xi), where h(x) is the hash function. In this problem we will assume, that h(x) = x mod p. Operation a mod b denotes taking a remainder after division a 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 the i-th insertion, you should output i. If no conflict happens, just output -1.

Input

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

Output

Output a single integer — the answer to the problem.

Sample test(s)

input
10 5
0
21
53
41
53

output
4

input
5 5
0
1
2
3
4

output
-1
求n 个数对同一个数取余  有余数相同的则输出在第多少个数遇到的相同的数  要求输出最小的
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    __int64 n,m,i,a[305],x,l,j,mixn;
    int k;
    while(cin>>n>>m)
    {
        k=0;
        mixn=1000000000;
        for(i=1;i<=m;i++)
        {
            cin>>x;
            a[i]=x%n;
        }
        /*for(i=1;i<=m;i++)
        cout<<a[i]<<" ";*/
        for( j=1;j<=m;j++)
          {
              for(i=j+1;i<=m;i++)
               {
                     if(a[j]==a[i])
                      {
                         l=i;
                         if(mixn>l)
                         mixn=l;
                         k=1;
                       }
                }
          }
             if(k)
        cout<<mixn<<endl;
        else
            cout<<"-1"<<endl;
    }
    return 0;
}