标签：des io os ar for div sp cti on
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 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 <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; }
Codeforces Round #267 (Div. 2) B. Fedor and New Game
标签：des io os ar for div sp cti on
原文地址：http://blog.csdn.net/u011345136/article/details/39394719
