Codeforces Round #267 (Div. 2) B. Fedor and New Game
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.
InputThe 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.
OutputPrint a single integer — the number of Fedor's potential friends.
Sample test(s) Input7 3 1 8 5 111 17Output
0Input
3 3 3 1 2 3 4Output
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#include#include #include #include 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<
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