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wordpress同步到本地,外贸网站seo教程,手机网站大全12345,网站建设 java水题。题目大意为输入一个p,n，再输入n组数据，每组数据对p取余，当和前面相同时，发现冲突，记下第几组，找到最后一组没有找到和前面一样的输出-1。 *可以采用一个标记数组，就变得很简单了。 A. DZ…

水题。题目大意为输入一个p,n，再输入n组数据，每组数据对p取余，当和前面相同时，发现冲突，记下第几组，找到最后一组没有找到和前面一样的输出-1。

*可以采用一个标记数组，就变得很简单了。

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
#include<stdio.h>
#include<string.h>
int main()
{int p,n;while(scanf("%d%d",&p,&n)!=EOF){int i;int x[305];int t;int tag=1;int temp;int w[304];//标记数组memset(w,0,sizeof(w));for(i=0;i<n;i++)scanf("%d",&x[i]);for(i=0;i<n;i++){t=x[i]%p;if(w[t]==0)w[t]=1;//进行标记else{tag=0;temp=i;break;}}if(tag==0)printf("%d\n",temp+1);elseprintf("-1\n");}return 0;
}
 
 