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

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.

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.

Print a single integer — the number of Fedor‘s potential friends.

7 3 1
8
5
111
17

0

3 3 3
1
2
3
4

3

【分析】：

“<< ”   位左移，相當於乘以2
 “>>"   位右移，相當於除以2
 “&”    按為與，位數都為1時為1，否則為0
 “^”    異或，不同為1，相同為0


新技能： a & (1 << j )  表示數a二進制的第j位是什麽

【代碼】：

#include <bits/stdc++.h>

using namespace std;
const int maxn = 1005;
int n,m,k,a[maxn];

int main()
{
    int sum=0,ans=0;
    cin>>n>>m>>k;
    for(int i=0;i<=m;i++)
        cin>>a[i];
    for(int i=0;i<m;i++)
    {
        sum=0; //註意內部清零
        for(int j=0;j<n;j++)
           if( ( a[i]&(1<<j) )^( a[m]&(1<<j) ) ) sum++;
        if(sum<=k) ans++;
    }

    cout<<ans<<endl;
    return 0;
}

Codeforces Round #267 (Div. 2) B. Fedor and New Game【位運算/給你m+1個數讓你判斷所給數的二進制形式與第m+1個數不相同的位數是不是小於等於k，是的話就累計起來】