Codeforces Round #267 (Div. 2) B. Fedor and New Game【位运算/给你m+1个数让你判断所给数的二进制形式与第m+1个数不相同的位数是不是小于等于k,是的话就累计起来】
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.
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,相当于乘以2
“>>" 位右移,相当于除以2
“&” 按为与,位数都为1时为1,否则为0
“^” 异或,不同为1,相同为0 新技能: a & ( << j ) 表示数a二进制的第j位是什么
【代码】:
#include <bits/stdc++.h> using namespace std;
const int maxn = ;
int n,m,k,a[maxn]; int main()
{
int sum=,ans=;
cin>>n>>m>>k;
for(int i=;i<=m;i++)
cin>>a[i];
for(int i=;i<m;i++)
{
sum=; //注意内部清零
for(int j=;j<n;j++)
if( ( a[i]&(<<j) )^( a[m]&(<<j) ) ) sum++;
if(sum<=k) ans++;
} cout<<ans<<endl;
return ;
}
Codeforces Round #267 (Div. 2) B. Fedor and New Game【位运算/给你m+1个数让你判断所给数的二进制形式与第m+1个数不相同的位数是不是小于等于k,是的话就累计起来】的更多相关文章
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