Codeforces Round #FF (Div. 二) A. DZY Loves Hash

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

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
题意：就是要你找出取模之后，找出相等的那一个下标

#include<string.h>
#include<stdio.h>
int main()
{   int  out=0;
__int64   p,n,i,j;
__int64   a[1001];
__int64  mod[1001];
while(scanf("%I64d %I64d",&p,&n)!=EOF)
{
	memset(a,0,sizeof(a));
	memset(mod,0,sizeof(mod));
	out=0;
	for(i=1;i<=n;i++)
	{
		scanf("%I64d",&a[i]);
		mod[i]=a[i]%p;
		
	}
/*		for(i=1;i<=n;i++)
	{
		printf("%d ",mod[i]);
		
	}
		printf("\n");
		*/
	for(i=2;i<=n&&out!=1;i++)
	{
		for(j=1;j<=i-1&&out!=1;j++)
		{
			//printf("%I64d-%I64d    ",i,j);
			if(mod[i]==mod[j])
			{
				

				printf("%I64d\n",i);
				out=1;
			}
		}
	//	printf("\n");
	}
	if(!out){
		printf("-1\n");
	}
}
return 0;
}

Codeforces Round #FF (Div. 二) A. DZY Loves Hash

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