CF447A DZY Loves Hash 模拟

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 x_i, DZY will put it into the bucket numbered h(x_i), 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 x_i (0 ≤ x_i ≤ 10⁹).

Output

Output a single integer — the answer to the problem.

Examples

Input

Copy

Output

Copy

Input

Copy

Output

Copy

-1
就是问hash是否存在冲突；
用 map 去判重即可；
当然开数组也没问题；

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstdlib>

#include<cstring>

#include<string>

#include<cmath>

#include<map>

#include<set>

#include<vector>

#include<queue>

#include<bitset>

#include<ctime>

#include<deque>

#include<stack>

#include<functional>

#include<sstream>

//#include<cctype>

//#pragma GCC optimize(2)

using namespace std;

#define maxn 20005

#define inf 0x7fffffff

//#define INF 1e18

#define rdint(x) scanf("%d",&x)

#define rdllt(x) scanf("%lld",&x)

#define rdult(x) scanf("%lu",&x)

#define rdlf(x) scanf("%lf",&x)

#define rdstr(x) scanf("%s",x)

typedef long long  ll;

typedef unsigned long long ull;

typedef unsigned int U;

#define ms(x) memset((x),0,sizeof(x))

const long long int mod = 1e9 + 7;

#define Mod 1000000000

#define sq(x) (x)*(x)

#define eps 1e-3

typedef pair<int, int> pii;

#define pi acos(-1.0)

//const int N = 1005;

#define REP(i,n) for(int i=0;i<(n);i++)

typedef pair<int, int> pii;

inline ll rd() {

    ll x = 0;

    char c = getchar();

    bool f = false;

    while (!isdigit(c)) {

        if (c == '-') f = true;

        c = getchar();

    }

    while (isdigit(c)) {

        x = (x << 1) + (x << 3) + (c ^ 48);

        c = getchar();

    }

    return f ? -x : x;

}

ll gcd(ll a, ll b) {

    return b == 0 ? a : gcd(b, a%b);

}

ll sqr(ll x) { return x * x; }

/*ll ans;

ll exgcd(ll a, ll b, ll &x, ll &y) {

    if (!b) {

        x = 1; y = 0; return a;

    }

    ans = exgcd(b, a%b, x, y);

    ll t = x; x = y; y = t - a / b * y;

    return ans;

}

*/

int n, p;

int x[maxn];

map<int, int>mp;

int main() {

    //ios::sync_with_stdio(0);

    cin >> p >> n;

    int pos = -1;

    for (int i = 1; i <= n; i++)cin >> x[i];

    for (int i = 1; i <= n; i++) {

        int tmp = x[i];

        if (!mp[tmp%p]) {

            mp[tmp%p] = 1;

        }

        else if (mp[tmp%p]) {

            pos = i; break;

        }

    }

    cout << pos << endl;

    return 0;

}

巴特西

CF447A DZY Loves Hash 模拟

最新文章

热门文章