# [Codeforces447A]DZY Loves Hash

## 描述 Description

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.

## 输入格式 InputFormat

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).

## 输出格式 OutputFormat

Output a single integer — the answer to the problem.

10 5
0
21
53
41
53

4

Codeforces 447A

## 代码 Code

``````#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf=0x7fffffff;
bool b[305];
int i,j,t,n,m,l,r,k,z,y,x,p,ans;
int main()
{
memset(b,false,sizeof(b));
ans=-1;
scanf("%d%d",&p,&n);
for (i=1;i<=n;i++)
{
scanf("%d",&x);
if (b[x%p] && ans==-1) ans=i;
b[x%p]=true;
}
printf("%d\n",ans);
return 0;
}``````