标签：codeforces 算法 algorithm acm
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
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
typedef long long LL;
int a[1005];
int main()
{
int n, m, k;
int mask, mm, ans;
cin >> n >> m >> k;
mask = (1<<n)-1;
REP(i, 1, m+1)
cin >> a[i];
mm = a[m+1];
ans = 0;
REP(i, 1, m)
{
int j = (a[i]^mm)&mask;
int cnt = 0;
REP(k, 0, n-1)
if ((j>>k)&1)
++cnt;
if (cnt <= k)
++ans;
}
cout << ans << endl;
return 0;
}
Codeforces Round #267 (Div. 2) B
标签：codeforces 算法 algorithm acm
原文地址：http://blog.csdn.net/notdeep__acm/article/details/39388331
