Fedor and New Game_u - fedor and new game-优快云博客

本文探讨了在一款名为《Call of Soldiers 3》的游戏中，玩家如何通过军队配置来寻找最多的朋友，涉及军队配置差异度的计算及二进制数的运算原理。

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Description

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 x_i. Consider binary representation of x_i: if the j-th bit of number x_i 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 x_i(1 ≤ x_i ≤ 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 Input

Input

Output

Input

Output

x&(1<<j) 表示数a二进制的第j位是什么

<pre name="code" class="cpp">(x[i]&(1<<j))^(x[m+1]&(1<<j)) 异或比较两数2进制的第j位，相同为0，相异为1

#include<stdio.h>
#include<algorithm>
using namespace std;

int main(){
    int n,m,k,x[1010],i,ans,j;
    ans=0;
    scanf("%d%d%d",&n,&m,&k);
    for(i=1;i<=m+1;i++){
        scanf("%d",&x[i]);
    }
    for(i=1;i<=m;i++){
        int temp=0;
        for(j=0;j<n;j++){
            if((x[i]&(1<<j))^(x[m+1]&(1<<j))){
                temp++;
            }
        }
        if(temp<=k){
            ans++;
        }
    }
    printf("%d\n",ans);
    return 0;
}