Codeforces Round #267 (Div. 2) B. Fedor and New Game【位运算/给你m+1个数让你判断所给数的二进制形式与第m+1个数不相同的位数是不是小于等于k，是的话就累计起来】

After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3».

The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered form 1 to (m + 1). Types of soldiers are numbered from 0 to n - 1. Each player has an army. Army of the i-th player can be described by non-negative integer xi. Consider binary representation of xi: if the j-th bit of number xi equal to one, then the army of the i-th player has soldiers of the j-th type.

Fedor is the (m + 1)-th player of the game. He assume that two players can become friends if their armies differ in at most k types of soldiers (in other words, binary representations of the corresponding numbers differ in at most k bits). Help Fedor and count how many players can become his friends.

Input

The first line contains three integers n, m, k (1 ≤ k ≤ n ≤ 20; 1 ≤ m ≤ 1000).

The i-th of the next (m + 1) lines contains a single integer xi (1 ≤ xi ≤ 2n - 1), that describes the i-th player's army. We remind you that Fedor is the (m + 1)-th player.

Output

Print a single integer — the number of Fedor's potential friends.

Sample test(s)

Input

Output

Input

Output

3

【分析】：

“<< ”   位左移，相当于乘以2

 “>>"   位右移，相当于除以2

 “&”    按为与，位数都为1时为1，否则为0

 “^”    异或，不同为1，相同为0

新技能： a & ( << j )  表示数a二进制的第j位是什么

【代码】：

#include <bits/stdc++.h>

using namespace std;

const int maxn = ;

int n,m,k,a[maxn];

int main()

{

    int sum=,ans=;

    cin>>n>>m>>k;

    for(int i=;i<=m;i++)

        cin>>a[i];

    for(int i=;i<m;i++)

    {

        sum=; //注意内部清零

        for(int j=;j<n;j++)

           if( ( a[i]&(<<j) )^( a[m]&(<<j) ) ) sum++;

        if(sum<=k) ans++;

    }

    cout<<ans<<endl;

    return ;

}

巴特西

Codeforces Round #267 (Div. 2) B. Fedor and New Game【位运算/给你m+1个数让你判断所给数的二进制形式与第m+1个数不相同的位数是不是小于等于k，是的话就累计起来】

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