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
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 test(s)

Input

7 3 1
8
5
111
17

Output

0

Input

3 3 3
1
2
3
4

Output

3
题意：给你m+1个数让你推断所给的数的二进制形式与第m+1个数不想同的位数是否小于等于k。累计答案
思路：题意题
#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\n", ans);
	return 0;
}