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
//题意就是找相等的数，输出第二个的位置，但是要是最先发现的。
例如：10 5
1 2 2 2 1
输出是3而不是5，因为先找到2和2相等，如果只用for循环，找到的是1和1相等输出是5.
第4个样例卡了很久，没看懂题目。。。。
#include <iostream>
using namespace std;
int main()
{ __int64 a[400];
int n,t,i,j,p,k;
while(scanf("%d%d",&p,&n)!=EOF)
{ memset(a,0,sizeof(a));
t=0;
for(i=0;i<n;i++)
{
scanf("%I64d",&a[i]);
a[i]=a[i]%p;
}
k=n;
for(i=0;i<n-1;i++)
{
for(j=i+1;j<n;j++)
if(a[i]==a[j])
{
k=k<(j+1)?k:(j+1);
t=1;
}
}
if(t==1)
printf("%d\n",k);
if(t==0)
printf("-1\n");
}
return 0;
}
CF A. DZY Loves Hash,布布扣,bubuko.com
原文地址：http://blog.csdn.net/qq2256420822/article/details/37754289
