标签：views tps lower for standard style pre time ++
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，可是要是最先发现的。
比如：10 5
1 2 2 2 1
输出是3而不是5，由于先找到2和2相等，假设仅仅用for循环，找到的是1和1相等输出是5.
第4个例子卡了非常久，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; }
标签：views tps lower for standard style pre time ++
原文地址：https://www.cnblogs.com/ldxsuanfa/p/10800340.html
