大家好，又见面了，我是全栈君。
1 second
256 megabytes
standard input
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.
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 a single integer — the answer to the problem.
10 5 0 21 53 41 53
4
5 5 0 1 2 3 4
-1
题意就是找相等的数，输出第二个的位置，可是要是最先发现的。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath>
using namespace std;
#define f1(i, n) for(int i=0; i<n; i++)
#define f2(i, m) for(int i=1; i<=m; i++)
#define f3(i, n) for(int i=n; i>=0; i--)
#define M 1005
const int INF = 0x3f3f3f3f;
int main()
{
int p, n, m, i;
int ans[M];
memset(ans, 0, sizeof(ans));
scanf("%d%d",&p,&n);
for (i=1; i<=n; i++)
{
scanf("%d",&m);
if (ans[m%p]++ == 1)
break;
}
printf("%d\n",(i>n)?-1:i);
}
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