Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) 831C. Jury Marks【二分】【水题】

本篇讨论了一个电视节目中参与者初始得分的可能性问题。通过已知的评委打分和部分得分记录，利用算法找出参赛者起始得分的所有可能选项。

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).

Examples

input
4 1
-5 5 0 20
10

output
3

input
2 2
-2000 -2000
3998000 4000000

output
1

枚举出所有的情况，用二分去筛就好了

#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
#include<stdio.h>
#include<string>
#include<map>
#include<cstring>
#include<set>
#include<vector>
#include<iomanip>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn = 2010;

int n, k, a[maxn], b[maxn], num[maxn];
bool vis[maxn];

int find(int k)
{
	int l = 0, r = n, m = (l + r) >> 1;
	while (l <= r)
	{
		m = (l + r) >> 1;
		if (num[m] == k) return m;
		if (num[m] > k) r = m - 1;
		else l = m + 1;
	}
	return n;
}

int main()
{
	while (~scanf("%d%d", &n, &k))
	{
		for (int i = 0; i < n; i++)
		{
			scanf("%d", a + i);
			if (i == 0) num[0] = a[0];
			else
			{
				num[i] = num[i - 1] + a[i];
			}
		}
		sort(num, num + n);
		for (int i = 0; i < k; i++) scanf("%d", b + i);
		sort(b, b + k);
		set<int> se;
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < k; j++)
			{
				se.insert(b[j] - num[i]);
			}
		}
		set<int>::iterator it;
		int ans = 0;
		for (it = se.begin(); it != se.end(); it++)
		{
			bool ok = 1;
			for (int i = 0; i < k; i++)
			{
				if (find(b[i] - *it) == n)
				{
					ok = 0;
					break;
				}
			}
			if (ok) ans++;
		}
		printf("%d\n", ans);
	}
}