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CF A. DZY Loves Hash

2024-07-11 23:00:33   229人阅读

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 thei-th number xi, DZY will put it into the bucket numberedh(xi), whereh(x) is the hash function. In this problem we will assume, thath(x)?=?x mod p. Operationa mod b denotes taking a remainder after divisiona 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 thei-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. Thei-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

//题意就是找相等的数，输出第二个的位置，但是要是最先发现的。例如：10 5      1 2 2 2 1输出是3而不是5，因为先找到2和2相等，如果只用for循环，找到的是1和1相等输出是5.第4个样例卡了很久，没看懂题目。。。。
#include <iostream>
using namespace std;
int main()
{   __int64 a[400];
    int n,t,i,j,p,k;
    while(scanf("%d%d",&p,&n)!=EOF)
    {   memset(a,0,sizeof(a));
        t=0;
        for(i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
            a[i]=a[i]%p;
        }
        k=n;
        for(i=0;i<n-1;i++)
        {    
            for(j=i+1;j<n;j++)
                if(a[i]==a[j])
                {   
                    k=k<(j+1)?k:(j+1);
                        t=1;
                }
                
        }
        if(t==1)
            printf("%d\n",k);
       if(t==0)
           printf("-1\n");
    }
    return 0;
}


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