第十二届浙江省大学生程序设计大赛-Team Formation 分类： ...

最新推荐文章于 2023-04-17 17:08:44 发布

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原文链接：http://www.cnblogs.com/juechen/p/4721980.html

教育者埃德华正在为其大学组建一支技能匹配的两成员队伍。通过位运算的异或操作来计算队伍的总技能水平，并判断是否大于队伍成员的个体技能水平，以决定队伍是否能够表现出色。

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Team Formation
Time Limit: 3 Seconds Memory Limit: 131072 KB

For an upcoming programming contest, Edward, the headmaster of Marjar University, is forming a two-man team from N students of his university.

Edward knows the skill level of each student. He has found that if two students with skill level A and B form a team, the skill level of the team will be A ⊕ B, where ⊕ means bitwise exclusive or. A team will play well if and only if the skill level of the team is greater than the skill level of each team member (i.e. A ⊕ B > max{A, B}).

Edward wants to form a team that will play well in the contest. Please tell him the possible number of such teams. Two teams are considered different if there is at least one different team member.
Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (2 <= N <= 100000), which indicates the number of student. The next line contains N positive integers separated by spaces. The ith integer denotes the skill level of ith student. Every integer will not exceed 109.
Output

For each case, print the answer in one line.
Sample Input

2
3
1 2 3
5
1 2 3 4 5

Sample Output

1
6

#include <iostream>
#include <string>
#include <stdio.h>
#include<algorithm>
#include<string.h>
#include<cmath>
#include<stack>
#include<queue>

using namespace std;

const int MAX=100000+100;

int a[MAX],bit[35];

void solve(int x)
{
    int b=31;

    while(b>=0)
    {
        if(x&(1<<b))
        {
            bit[b]++;

            return ;

        }
        b--;
    }
}

int main()
{

    int T,n,i,sum;

    scanf("%d",&T);

    while(T--)
    {

        sum=0;

        memset(bit,0,sizeof(bit));

        scanf("%d",&n);

        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);

            solve(a[i]);

        }

        for(i=0;i<n;i++)
        {
            int b=31;

            while(b>=0)
            {
                if(a[i]&(1<<b))
                {
                    break;
                }

                b--;

            }

            while(b>=0)
            {
                if(!(a[i]&(1<<b)))
                {
                    sum+=bit[b];
                }
                b--;
            }

        }
        printf("%d\n",sum);

    }

    return 0;
}