[CodeForces - 447A] A - DZY Loves Hash

最新推荐文章于 2019-01-03 16:48:00 发布

转载最新推荐文章于 2019-01-03 16:48:00 发布 · 270 阅读

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原文链接：http://www.cnblogs.com/whc200305/p/7215755.html

本文介绍了一种简单的Hash冲突检测算法，通过模拟插入操作来判断从哪个操作开始出现Hash值重复，解决基本的数据冲突问题。文章提供了一个C++实现示例，展示了如何使用数组记录已插入的Hash值，并在发生冲突时返回当前操作序号。

A - DZY Loves Hash

DZY has a hash table with p buckets, numbered from 0 to p - 1. He wants to insert nnumbers, 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.

Example

Input

Output

Input

Output

-1

题目的意思是,按顺序给出几个数,判断从哪个操作开始,hash值有重复.

题目很水,模拟一下就好了.

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 bool vis[1000];
 6 int main(){
 7     int n,m;
 8     scanf("%d%d",&m,&n);
 9     for (int i=1; i<=n; i++){
10         int x; scanf("%d",&x);
11         x%=m; if (vis[x]){printf("%d",i); return 0;}
12         vis[x]=1;
13     }
14     puts("-1");
15     return 0;
16 }

View Code

转载于:https://www.cnblogs.com/whc200305/p/7215755.html