CodeForces 831C : Jury Marks（set的使用）

Jury Marks

Time limit：2000 ms Memory limit：262144 kB

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Problem Description

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 ai 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 k judges rated the participant there were n (n ≤ k) values b1, b2, …, bn (it is guaranteed that all values bj 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, …, ak ( - 2 000 ≤ ai ≤ 2 000) — jury’s marks in chronological order.

The third line contains n distinct integers b1, b2, …, bn ( - 4 000 000 ≤ bj ≤ 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).

Example

Note

The answer for the first example is 3 because initially the participant could have  - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.

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题意：

Polycarp有一个初始分数，k个评委按照顺序打分（加分或者扣分），评委打分后分数变化，得到一个中间的分数，但是Polycarp只记得一部分中间分数了（n个），问所有初始可能的分数有几种。

解题思路：

暴力。把评委打分的所有可能（1,1~2，1~3…1~n）的和加入到set里，然后再用中间分数减去这个数，就可以得到可能的初始分数，这种可能的初始分数要出现n次,也就是说对所有的中间值都有效，才能算真正可能的初始分数。统计次数的方法应该有很多，我这里用的是map，结果差点超时（1902/2000）。。

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Code：

#include <iostream>#include <cstring>#include <set>#include <map>using namespace std;const int M=2000+5;int mark[M];int sc[M];int sum[M];set<int> s;map<int,int> m;set<int>::iterator sit;map<int,int>::iterator mit;int main(){    int n,k;    while(cin>>k>>n)    {         s.clear();         m.clear();        for(int i=0;i<k;i++)            cin>>mark[i];        for(int j=0;j<n;j++)            cin>>sc[j];        memset(sum,0,sizeof(sum));        sum[0]=mark[0];        s.insert(sum[0]);        for(int i=1;i<k;i++)        {            sum[i]=sum[i-1]+mark[i];            s.insert(sum[i]);        }        for(int i=0;i<n;i++)        {            for(sit=s.begin();sit!=s.end();sit++)            {                m[sc[i]-*sit]++;                //cout<<sc[i]-*sit<<endl;            }        }        int ans=0;        for(mit=m.begin();mit!=m.end();mit++)        {            if((*mit).second==n)            {                ans++;                //cout<<(*mit).first<<endl;            }        }        cout<<ans<<endl;    }    return 0;}