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

A. DZY Loves Hash

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 x_i, DZY will put it into the bucket numbered
h(x_i), 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 x_i
(0 ≤ x_i ≤ 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

题意就是找相等的数，输出第二个的位置，可是要是最先发现的。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath>

using namespace std;

#define f1(i, n) for(int i=0; i<n; i++)
#define f2(i, m) for(int i=1; i<=m; i++)
#define f3(i, n) for(int i=n; i>=0; i--)
#define M 1005

const int INF = 0x3f3f3f3f;

int main()
{

    int p, n, m, i;
    int ans[M];
    memset(ans, 0, sizeof(ans));
    scanf("%d%d",&p,&n);
    for (i=1; i<=n; i++)
    {
        scanf("%d",&m);
        if (ans[m%p]++ == 1)
            break;
    }
    printf("%d\n",(i>n)?-1:i);

}