Codeforces 789E The Great Mixing （数推倒公式 + bfs + 剪枝）

题意：有k个数,是1/1000的倍数，问能否选取任意个数，每个数也能选任意次，使他们的均值为n/1000

题解:假设选了m个数 (s1+s2+...sm)/1000/m=n/1000,化简后得到(s1-n)+(s2-n)+....+(sm-n)=0,原题转化成从k个数中选m个数使他们的和为0，且选出来的数的和的范围必然是在[-1000,1000],这样用bfs来写，就可以得到最少需要选几个数，O(2001*min(k,1001))

题目就转换成：有1个序列， 每个数字可以选多次，问最少取几个数字能组成0。。。。

那么就是一个简单的搜索题，注意一下剪枝就行啦~~~~~

Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration . Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration . The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.

Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.

Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration . Assume that the friends have unlimited amount of each Coke type.

Input

The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.

The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.

Output

Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration , or -1 if it is impossible.

Examples

input

400 4100 300 450 500

output

2

input

50 2100 25

output

3

Note

In the first sample case, we can achieve concentration  using one liter of Coke of types  and : .

In the second case, we can achieve concentration  using two liters of  type and one liter of  type: .

#include<bits/stdc++.h>using namespace std;map<int,int>vis,vis1;const int N = 1e6 + 100;const int inf = 0x3f3f3f3f;struct P{    int x,step;    P(int x,int step):x(x),step(step){}    P(){}};int a[N];int bfs(){    queue<P>que;    que.push(P(0,0));    int i,j;    while(!que.empty()){        P now = que.front();    //    cout<<now.x<<endl;        que.pop();        for(i=-1000;i<=1000;i++) {            if(vis[i]) {                if(now.x>0 && i>0) continue;                if(now.x<0 && i<0) continue;                if(now.x+i==0) return now.step+1;                if(vis1[now.x+i]) continue;                vis1[now.x+i]=1;                que.push(P(now.x+i,now.step+1));            }        }    }}int main(){    ios::sync_with_stdio(false);    int n,k,i,Min,Max,zero;    cin>>n>>k;    Min=inf,Max=-inf,zero=0;    for(i=1;i<=k;i++) {        cin>>a[i];        a[i]-=n;        if(a[i]==0) zero=1;        vis[a[i]]=1;        Min=min(Min,a[i]);        Max=max(Max,a[i]);    }    if(zero) {        cout<<"1"<<endl;    }    else if(Min*Max>0) {        cout<<"-1"<<endl;    }    else {        int ans = bfs();        cout<<ans<<endl;    }    return 0;}