Codeforces Round #267 (Div. 2) B. Fedor and New Game【位運算/給你m+1個數讓你判斷所給數的二進制形式與第m+1個數不相同的位數是不是小於等於k,是的話就累計起來】
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
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.
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
【分析】:
“<< ” 位左移,相當於乘以2 “>>" 位右移,相當於除以2 “&” 按為與,位數都為1時為1,否則為0 “^” 異或,不同為1,相同為0 新技能: a & (1 << j ) 表示數a二進制的第j位是什麽
【代碼】:
#include <bits/stdc++.h> using namespace std; const int maxn = 1005; int n,m,k,a[maxn]; int main() { int sum=0,ans=0; cin>>n>>m>>k; for(int i=0;i<=m;i++) cin>>a[i]; for(int i=0;i<m;i++) { sum=0; //註意內部清零 for(int j=0;j<n;j++) if( ( a[i]&(1<<j) )^( a[m]&(1<<j) ) ) sum++; if(sum<=k) ans++; } cout<<ans<<endl; return 0; }View Code
Codeforces Round #267 (Div. 2) B. Fedor and New Game【位運算/給你m+1個數讓你判斷所給數的二進制形式與第m+1個數不相同的位數是不是小於等於k,是的話就累計起來】