cf-467B. Fedor and New Game

467B. Fedor and New Game

                        time limit per test 1 second  memory limit per test 256 megabytes
                                input standard input  output standard output

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.

题目思路：就是分别与Fedor异或，求结果中1的个数。可以利用__builtin_popcount（）函数：用于计算二进制中含有多少个1.

题目链接：codeforces 467B

以下是代码：

#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
using namespace std;

int main(){
    int n,m,k;
    cin >> n >> m >> k;
    int num[2000] = {0};
    for (int i = 0; i < m; i++)
    {
        cin >> num[i];
    }
    int me;
    cin >> me;
    int res = 0;
    for (int i = 0; i < m; i++)
    {
        res += __builtin_popcount(me ^ num[i]) <= k; 
    }
    cout << res << endl;
    return 0;
}