Codeforces Round #FF (Div. 2):Problem A - DZY Loves Hash
2014-07-14 08:07
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A. DZY Loves Hash
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Input
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
Output a single integer — the answer to the problem.
Sample test(s)
Input
Output
Input
Output
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Input
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
Output a single integer — the answer to the problem.
Sample test(s)
Input
10 5 0 21 53 41 53
Output
4
Input
5 5
0
1
2
3
4
Output
-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。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> #include<queue> #include<sstream> #include<cmath> using namespace std; #define f1(i, n) for(int i=0; i<n; i++) #define f2(i, m) for(int i=1; i<=m; i++) #define f3(i, n) for(int i=n; i>=0; i--) #define M 1005 const int INF = 0x3f3f3f3f; int main() { int p, n, m, i; int ans[M]; memset(ans, 0, sizeof(ans)); scanf("%d%d",&p,&n); for (i=1; i<=n; i++) { scanf("%d",&m); if (ans[m%p]++ == 1) break; } printf("%d\n",(i>n)?-1:i); }
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