标签：art 条件 insert div spec scanf ems nes eof
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
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;
}
Codeforces Round #FF (Div. 2) A. DZY Loves Hash
标签：art 条件 insert div spec scanf ems nes eof
原文地址：http://www.cnblogs.com/lxjshuju/p/7234827.html
