CF 447A(DZY Loves Hash-简单判重)

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 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

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;
}