Codeforces Round #FF (Div. 二/A)/Codeforces447A_DZY Loves Hash(哈希)

Codeforces Round #FF (Div. 2/A)/Codeforces447A_DZY Loves Hash(哈希)

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

解题报告

水，好像写麻烦了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
int p,n,i,num[1000],j;
vector<int>hashh[1000];
int main()
{

    while(cin>>p>>n)
    {
        memset(hashh,0,sizeof(hashh));
        for(i=0;i<n;i++)
        {
            cin>>num[i];
        }
        for(i=0;i<n;i++)
        {
            int sz=hashh[num[i]%p].size();
            if(sz)
            {cout<<i+1<<endl;
            break;}
            else hashh[num[i]%p].push_back(num[i]);
        }
        if(i==n)cout<<"-1"<<endl;
    }
    return 0;
}

Codeforces Round #FF (Div. 二/A)/Codeforces447A_DZY Loves Hash(哈希)

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