CodeForces 831C Jury Marks（set）

最新推荐文章于 2018-04-02 20:41:53 发布

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最新推荐文章于 2018-04-02 20:41:53 发布

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分类专栏： CodeForces 【C++STL】

本文链接：https://blog.youkuaiyun.com/qq_34845082/article/details/75116622

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本文介绍了一种通过已知部分比赛过程得分来推测参赛者初始得分的方法。文章详细阐述了如何利用前缀和与集合操作解决此类问题，并提供了一个具体的算法实现案例。

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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 thek judges rated the participant there weren (n ≤ k) valuesb₁, 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 andn (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).

Examples

Input

4 1
-5 5 0 20
10

Output

Input

2 2
-2000 -2000
3998000 4000000

Output

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.

【题意】一个人有一个初始积分，接下来有k个人依次为他加分，然后告诉你n(1<=n<=k)个过程中的积分，问他的初始积分有多少可能。

【思路】首先根据每次的加分做一个前缀和运算，来获得到这个点为止所加的分，此时前缀和里不同值的个数就是答案的最大值。枚举初始积分可能的值，并判断在这个前提下所有n个过程中的积分是否都出现过。这里用set的size()和count()来实现会很方便~

#include<iostream>
#include<cstdio>
#include<set>
#include<algorithm>
using namespace std;
const int MAXN=2005;
int a[MAXN],b[MAXN],c[MAXN];
int n,k;
set<int> s;
set<int>::iterator its;

int main(){
    int i,j;
    int e,ans,x;
    while(cin>>k>>n){
        b[1]=0;
        s.clear();
        for(i=1;i<=k;i++){
            scanf("%d",&a[i]);
            b[i]+=a[i];
            b[i+1]=b[i];
            s.insert(b[i]);
        }
        ans=s.size();
        for(i=1;i<=n;i++){
            scanf("%d",&c[i]);
        }
        for (its=s.begin();its!=s.end();its++){
            x=*its;
            e=c[1]-x;
            for(j=1;j<=n;j++){
                if(!s.count(c[j]-e)){
                    ans--;
                    break;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

【总结】昨天没做出来。其实思路也就是那样…没做出来主要是，感觉当时思维挺混乱的，想的是一个过程分可能出现在所有不同时间点上，要先枚举所有可能的出现点，然后根据这个过程分和其他过程分的差值对应着前缀和之间的差值来一一匹配到所有时间点上…？？好像完全忘记了初始分这个绝对判断点的存在。