12th浙江省省赛 Team Formation

最新推荐文章于 2019-04-16 21:34:09 发布

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最新推荐文章于 2019-04-16 21:34:09 发布

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分类专栏：算法基础文章标签： acm省赛

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本文探讨了一个算法问题，即如何从N名学生中选择两名技能水平合适的成员组成一支参赛队伍。队伍的技能水平由成员技能水平的按位异或运算决定，并且只有当队伍的技能水平高于每个成员的个人技能水平时，这支队伍才能发挥出色。文章提供了一种解决方案，通过数位dp或寻找规律的方法来计算可能的优秀队伍数量。

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 i^th integer denotes the skill level of i^th student. Every integer will not exceed 10⁹.

Output

For each case, print the answer in one line.

Sample Input

Sample Output

1
6
思路：数位dp 或者 找规律：）
把异或表打出来 找数与数之间的关系
 #include<bits/stdc++.h>
using namespace std;
int num[100020];
int sum[100];
int main()
{
    //freopen("b.txt","r",stdin);
    int t;
    cin>>t;
    while(t--)
    {
        memset(sum,0,sizeof(sum));
        int n;
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%d",&num[i]);
        sort(num,num+n);
        int ans=0;
        for(int i=0; i<n; i++)
        {
            int a=num[i];
            int wei=0;
            while(a)
            {
                a/=2;
                wei++;
            }
            wei--;
            //cout<<"--->"<<wei<<endl;
            int k=1;
            for(int j=0; j<wei; j++)
            {
                int chu=num[i]/k;
                if(chu%2==0)
                    ans+=sum[j];
                //cout<<i<<"   "<<j<<"  "<<sum[j]<<"  "<<ans<<endl;
                k*=2;
            }
            //cout<<"1."<<i<<"   "<<wei<<"  "<<sum[wei]<<endl;
            sum[wei]++;
            //cout<<"2."<<i<<"   "<<wei<<"  "<<sum[wei]<<endl;
        }
        cout<<ans<<endl;
    }
    return 0;
}