Codeforces Round #267 (Div. 2) B. Fedor and New Game

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

7 3 1
8
5
111
17

Output

0

Input

3 3 3
1
2
3
4

Output

3
  ：  m+1                  m+1              k，    
  ：   
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1010;

int num[maxn], cnt;

int main() {
	int n, m, k;
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 0; i < m; i++)
		scanf("%d", &num[i]);
	scanf("%d", &cnt);
	int ans = 0;
	for (int i = 0; i < m; i++) {
		int cur = 0;
		for (int j = 0; j < n; j++)
			if (((1<<j)&num[i]) != ((1<<j)&cnt))
				cur++;
		if (cur <= k)
			ans++;
	}
	printf("%d
", ans);
	return 0;
}