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, DZY will put it into the bucket numbered
h(x), 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 x
(0 ≤ x ≤ 10).
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
注意数据范围p,m<=200。尽管输入的数据较大，刚開始自己mod1000000001，发现这不能有冲突了，而且还要标记下第一个发生冲突，以后的冲突能够忽略
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
using namespace std;
int ha[10002];
int bj;
int main()
{
LL n,m,i,j,k,l;
while(~scanf("%d%d",&n,&m))
{
bj=0;
memset(ha,0,sizeof(ha) );
for(i=1;i<=m;i++)
{
scanf("%d",&k);
k=k%n;//对当前的输入值的个数进行%
if(!ha[k])
{
ha[k]++;
}
else if(ha[k]!=0&&!bj )//仅仅取。符合条件的第一个<span id="transmark"></span>
{
bj=i;
}
}
if(!bj)
printf("-1\n");
else printf("%d\n",bj);
}
return 0;
}