【codeforces 467B】 Fedor and New Game(位运算模拟)

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本文介绍了一种通过计算玩家军队中士兵类型的差异来确定潜在朋友数量的方法。利用位运算技巧，特别是异或操作，可以有效地解决这个问题。文章还提供了一个C++实现示例。

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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.

Examples
inputCopy
7 3 1
8
5
111
17
outputCopy
0
inputCopy
3 3 3
1
2
3
4
outputCopy
3

直接用异或进行简单的位运算，计算过后再来判断运算后1的个数是否超过了k，若没有，则ans++；

#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 10005
int a[MAXN];
int main()
{
	int n,m,k;
	cin>>n>>m>>k;
	int play;
	for(int i=1;i<=m;i++)
	{
		cin>>a[i];
	}
	cin>>play;
	int ans=0;
	for(int i=1;i<=m;i++)
	{
		int t=0;
		int p=a[i]^play;
		while(p)
		{
			if(p&1) t++;
			p>>=1;
		}
		if(t<=k) ans++;
	}
	cout<<ans<<endl;
	return 0;
}