CodeForces 447-A. DZY Loves Hash
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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 5021534153
4
5 501234
-1
#include<stdio.h>#include<iostream>#include<string.h>using namespace std;int main(){ int a[310],num[310]; memset(num,0,sizeof(num)); int p,n; scanf("%d%d",&p,&n); for(int i = 0; i < n;i++) { scanf("%d",&a[i]); } int flag = 0; int ans = 0; for(int i = 0;i < n;i++) { if(num[a[i]%p] == 0) num[a[i]%p]++; else { flag = 1; ans = i+1; break; } } if(flag) printf("%d\n",ans); else printf("-1\n"); return 0;}
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