1 second
256 megabytes
standard input
standard output
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
sb题*2
给m个数，转成二进制，问和给定的x的每一位异或和小等于k的有多少个
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m,k,ans,a[100100];
int work(int x)
{
int res=0;
for(int i=0;i<=21;i++)
if((x&(1<<i))^(m&(1<<i)))res++;
return res;
}
int main()
{
scanf("%d%d%d",&m,&n,&k);
int x,y;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
scanf("%d",&m);
for(int i=1;i<=n;i++)
if(work(a[i])<=k)ans++;
printf("%d",ans);
return 0;
}