Jury Marks

Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave a_i points.

Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the kjudges rated the participant there were n (n ≤ k) values b₁, b₂, ..., b_n (it is guaranteed that all values b_j are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.

Your task is to determine the number of options for the score the participant could have before the judges rated the participant.

Input

The first line contains two integers k and n (1 ≤ n ≤ k ≤ 2 000) — the number of jury members and the number of scores Polycarp remembers.

The second line contains k integers a₁, a₂, ..., a_k ( - 2 000 ≤ a_i ≤ 2 000) — jury's marks in chronological order.

The third line contains n distinct integers b₁, b₂, ..., b_n ( - 4 000 000 ≤ b_j ≤ 4 000 000) — the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.

Output

Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).

利用vector数据结构，利用去重函数，

#include<iostream>#include<vector>#include<algorithm>#include<map>#include<cstdio>using namespace std;vector <int> v;map <int,int> m;int sum[2005],k,n,x,ans=0;int main(){    sum[0]=0;    scanf("%d%d",&n,&k);    for(int i=1;i<=n;i++)    {        scanf("%d",&x);        sum[i]=sum[i-1]+x;    }    sort(sum+1,sum+1+n);    int len=unique(sum+1,sum+1+n)-(sum+1);    for(int i=0;i<k;i++)    {        scanf("%d",&x);        for(int j=1;j<=len;j++) v.push_back(x-sum[j]);    }    sort(v.begin(),v.end());    for (int i=0;i<v.size();i++) m[v[i]]++;    for (int i=0;i<v.size();i++)    {        if (m[v[i]]==k)        {            ans++;            m[v[i]]=0;        }    }    printf("%d",ans);    return 0;}

一开始超时了了好久