Codeforces Round #356 (Div. 2)B. Bear and Finding Criminals（水题）

B. Bear and Finding Criminals

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 citya. 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 ti criminals in the i-th city.

Output

Print the number of criminals Limak will catch.

Examples

input

6 3
1 1 1 0 1 0

output

3

input

5 2
0 0 0 1 0

output

1

链接：http://codeforces.com/contest/680/problem/B
这题也是水题，直接看底下的NOTE，更容易懂。

#include<bits/stdc++.h>

using namespace std;

#define SI(N) scanf("%d",&(N))

int main()
{
    int a[105]={0};
    int n,in;
    cin>>n>>in;
    for (int i=1;i<=n;i++)
    {
        SI(a[i]);
    }
    int ans=0;
    int l=in-1,r=in+1;
    while(l>=1&&r<=n)
    {
        if (a[l]==1&&a[r]==1)
        {
            ans+=2;
        }
        l--,r++;
    }
    if (l==0)
    {
        for (int i=r;i<=n;i++)
        {
            if (a[i]==1)
                ans++;
        }
    }
    if (r==n+1)
    {
        for (int i=l;i>=0;i--)
        {
            if (a[i]==1)
                ans++;
        }
    }
    if (a[in]==1)
        ans++;
    printf("%d\n",ans);
    return 0;
}