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

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

                                                  

题意：

n 种士兵，m+1 个人（1<=m<=1000) ,求出前m个数和第m+1个数的二进制位不同的位数在k以内的个数。

新技能 a & (1 << j ) 表示数a二进制的第j 位是什么

CODE：

#include <iostream>
#include <stdio.h>
#include <string>
#include <algorithm>
#include <string.h>

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