二进制新知识（Codeforces Round #267 (Div. 2) B. Fedor and New Game）

                        J - Fedor and New Game

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status
Description
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 Input
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个数让你判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。
这里用到了一个二进制的知识： a & (1 << j ) 表示数a二进制的第j 位是什么。**

#include <bits/stdc++.h>#include <algorithm>using namespace std;int a[1010];int main(){    int n,m,k,x,ans=0;    scanf("%d%d%d",&n,&m,&k);    for(int i = 1; i <= m; i++)    {        scanf("%d",&a[i]);    }    scanf("%d",&x);    for(int i = 1; i <= m; i++)    {        int res=0;        for(int j = 0; j < n; j++)        {            if((a[i]&(1<<j)) != (x&(1<<j)))                res++;        }        if(res<=k)            ans++;    }    printf("%d\n",ans);    return 0;}