codeforces gym/100814 humming distance (二进制位数比较)

 Gym - 100814I 

I. Salem

time limit per test

1 second

memory limit per test

1024 megabytes

input

standard input

output

standard output

Salem is known to be one of the best competitive programmers in the region. However, he always finds a hard time understanding the concept of the hamming distance. The hamming distance of two numbers is defined as the number of different digits in their binary representations (leading zeros are used if necessary to make the binary representations have the same length). For example the hamming distance between 12 and 5 is 2 since their binary representations are 1100 and 0101 respectively and they differ in the first and fourth positions.

Recently, Salem got a problem that he needs your help to solve. He is given N integers and asked to get the maximum among the hamming distances of all possible pairs of the given integers.

Input

The first line of the input will be a single integer T representing the number of test cases. Followed by T test cases. Each test case will start with a line with single integer (2 ≤ N ≤ 100) representing the number of the integers. Each of the following N lines contains one of the integers (1 ≤ Ai ≤ 10, 000) from the list of the integers.

Output

For each test case print one line consists of one integer representing the maximum hamming distance between all possible pairs from the given integers.

Examples

input

2
2
12
5
3
1
2
3

output

2
2

参考：
http://codeforces.com/blog/entry/21567

两种方法：
1.

int humming(int x,int y)

2.

__builtin_popcount(arr[i]^arr[j]);/

#include "cstdio"
#include "iostream"
using namespace std;
int humming(int x,int y)
{
    int ans=0;
    while(1)
    {
        if(x==0&&y==0)break;
        if(x%2!=y%2)
        ans++;
        x/=2;y/=2;
    }
    return ans;
}
int main()
{
    int a,b;
    int t,n;
    int x,y;
    int arr[105];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&arr[i]);
        }
        int dist=0,max1=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i;j<n;j++)
            {
                //dist=__builtin_popcount(arr[i]^arr[j]);///求两数最大海明差距
                dist=humming(arr[i],arr[j]);
                max1=max1>dist?max1:dist;
            }
        }
        printf("%d
",max1);
    }
}