Codeforces Round #FF (Div. 2):Problem A

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 x_i, DZY will put it into the bucket numbered h(x_i), 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 x_i (0 ≤ x_i ≤ 10⁹).

Output

Output a single integer — the answer to the problem.

Sample test(s)

Input

Output

Input

Output

-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 1005const 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);}

转载于:https://www.cnblogs.com/yxwkf/p/5267254.html

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