CF 447A(DZY Loves Hash-简单判重)

A. DZY Loves Hash

time limit per test

memory limit per test

input

output

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.

i-th insertion, you should output i. If no conflict happens, just output -1.

Input

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

题目大意：有一堆数扔进Hash表，求第一次出现2个数在1个格子中的情况

模拟

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (300+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int p,n;
bool b[MAXN];
int main()
{
//  freopen("DZY Loves Hash.in","r",stdin);
//  freopen(".out","w",stdout);
  
  cin>>p>>n;
  MEM(b)
  For(i,n)
  {
    int x;
    cin>>x;
    x%=p;
    if (b[x]) {cout<<i<<endl; return 0;}
    else b[x]=1;
  }
  cout<<"-1"<<endl;
    
  
  
  return 0;
}