CUC-SUMMER-6-G

G - Bear and Finding Criminals

CodeForces-680B

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There are n cities in Bearland, numbered 1 through n. Cities are arranged in one long row. The distance between cities i and j is equal to |i - j|.
Limak is a police officer. He lives in a city a. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city a. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.

Input
The first line of the input contains two integers n and a (1 ≤ a ≤ n ≤ 100) — the number of cities and the index of city where Limak lives.
The second line contains n integers t1, t2, ..., tn(0 ≤ ti ≤ 1). There are ticriminals in the i-th city.

Output
Print the number of criminals Limak will catch.

Example
Input
6 31 1 1 0 1 0

Output
3

Input
5 20 0 0 1 0

Output
1

Note
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.

Using the BCD gives Limak the following information:
There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city.
There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city.
There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city.
There are zero criminals for every greater distance.

So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is.

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题意：在一维的坐标轴上，警察住在n位置，仪器会告诉你距离为k的地方有几个犯人，问警察最多能确定几个犯人

解法：从n向两侧遍历，如果两边都没到头能判断0和2，一端到头之后能判断0和1

代码：

#include<iostream>
using namespace std;
int c[105];
int main()
{
    int n,a;
    cin>>n>>a;
    for(int i=1;i<=n;i++)
        cin>>c[i];
    int k=1;
    int ans=c[a]?1:0;
    while(1){
        if(a-k>0&&a+k<=n){
            ans+=(c[a-k]&c[a+k])*2;
        }
        else if(a-k<1&&a+k<=n)
            ans+=c[a+k];
        else if(a-k>=1&&a+k>n)
            ans+=c[a-k];
        else
            break;
        k++;
    }
    cout<<ans<<endl;
    return 0;
}