Codeforces Round #267 (Div. 2) B. Fedor and New Game_html/css_WEB-ITnose

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发布: 2016-06-24 11:57:38
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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.

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Output

Print a single integer ? the number of Fedor's potential friends.

Sample test(s)

Input

7 3 18511117
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Output

Input

3 3 31234
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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,累计答案思路:题意题<pre name="code" class="n">#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1010;int num[maxn], cnt;int main() {	int n, m, k;	scanf("%d%d%d", &n, &m, &k);	for (int i = 0; i < m; i++)		scanf("%d", &num[i]);	scanf("%d", &cnt);	int ans = 0;	for (int i = 0; i < m; i++) {		int cur = 0;		for (int j = 0; j < n; j++)			if (((1<<j)&num[i]) != ((1<<j)&cnt))				cur++;		if (cur <= k)			ans++;	}	printf("%d\n", ans);	return 0;}
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