Codeforces Round #267 (Div. 2) B
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题目:
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 18511117
0
3 3 31234
3
题意分析:
代码:
#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;}
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