题干：
Vasya has the sequence consisting of n integers. Vasya consider the pair of integers x and y k-interesting, if their binary representation differs from each other exactly in k bits. For example, if k = 2, the pair of integers x = 5 and y = 3 is k-interesting, because their binary representation x=101 and y=011 differs exactly in two bits.
Vasya wants to know how many pairs of indexes (i, j) are in his sequence so that i < j and the pair of integers ai and aj is k-interesting. Your task is to help Vasya and determine this number.
Input
The first line contains two integers n and k (2 ≤ n ≤ 105, 0 ≤ k ≤ 14) — the number of integers in Vasya's sequence and the number of bits in which integers in k-interesting pair should differ.
The second line contains the sequence a1, a2, ..., an (0 ≤ ai ≤ 104), which Vasya has.
Output
Print the number of pairs (i, j) so that i < j and the pair of integers ai and aj is k-interesting.
Examples
Input
4 1
0 3 2 1
Output
4
Input
6 0
200 100 100 100 200 200
Output
6
Note
In the first test there are 4 k-interesting pairs:
- (1, 3),
- (1, 4),
- (2, 3),
- (2, 4).
In the second test k = 0. Consequently, integers in any k-interesting pair should be equal to themselves. Thus, for the second test there are 6 k-interesting pairs:
- (1, 5),
- (1, 6),
- (2, 3),
- (2, 4),
- (3, 4),
- (5, 6).
题目大意：
给你N个数的一个序列a，如果有两个数二进制中有K个位置不同，那么对应就算作一对可行解，问一共有多少对可行解。
(2 ≤ n ≤ 1e5, 0 ≤ k ≤ 14, 0 ≤ ai ≤ 1e4)
解题报告：
介绍这个题的两种写法。首先很显然看x和y有多少个位子不同，直接看x^y对应的二进制中1的个数即可。
所以第一种想法啊十分好想，预处理1~20000所有数字的1的个数，然后暴力枚举1e4以内的每个数字i和j，如果i^1的1的个数等于k，则直接统计i，j两个数的出现次数。我这里偷懒了没预处理，直接用内置函数了。这种方法复杂度O(1e8)
第二种方法就是读入完k之后，先暴力枚举所有1的个数等于k的数，打表可以发现对任意一个k，1e4以内不会超过3000个数。问题等价于求x^y=z，x和y符合的二元组个数，已知x<=1e4，z最多3000个，所以直接枚举x和z来反求y。复杂度O(3e8)左右。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 1<<15;
int n,k;
int dp[MAX];
vector<int> vv;
int cnt[MAX],mx;
ll ans;
int main()
{
// cout << (1<<14) << endl;cin >> n >> k;for(int x,i = 1; i<=n; i++) {scanf("%d",&x); dp[x]++; mx = max(mx,x);}if(k == 0) {for(int i = 0; i<=mx; i++) ans += 1LL*dp[i]*(dp[i]-1)/2;cout << ans << endl; return 0;}for(int i = 0; i<=mx; i++) {for(int j = i+1; j<=mx; j++) {if(__builtin_popcount(i^j) == k) {ans += 1LL*dp[i] * dp[j];}}} cout << ans << endl;return 0 ;
}
