Codeforces Round #FF (Div. 2) 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 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

求n 个数对同一个数取余  有余数相同的则输出在第多少个数遇到的相同的数  要求输出最小的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
__int64 n,m,i,a[305],x,l,j,mixn;
int k;
while(cin>>n>>m)
{
k=0;
mixn=1000000000;
for(i=1;i<=m;i++)
{
cin>>x;
a[i]=x%n;
}
/*for(i=1;i<=m;i++)
cout<<a[i]<<" ";*/
for( j=1;j<=m;j++)
{
for(i=j+1;i<=m;i++)
{
if(a[j]==a[i])
{
l=i;
if(mixn>l)
mixn=l;
k=1;
}
}
}
if(k)
cout<<mixn<<endl;
else
cout<<"-1"<<endl;
}
return 0;
}