Team Formation 浙江省省赛原题

最新推荐文章于 2020-10-07 16:18:11 发布

LiQiyaoo

最新推荐文章于 2020-10-07 16:18:11 发布

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分类专栏： acm 文章标签： acm Team Formation 位运算

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本文介绍了一个算法问题，探讨如何从N个学生中选择两名技能水平合适的成员组成参赛团队。问题的关键在于通过位运算判断两个成员技能相加后的总技能是否大于各自技能水平，以此来评估团队合作效果。

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B - Team Formation

Time Limit:3000MS Memory Limit:131072KB 64bit IO Format:%lld & %llu

Submit Status

Description

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

此题考察了很多位运算的知识。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int a[100005];
int b[50];//b[i]表示有多少个数转换为二进制后i是最高位 
void caculate(int num){
	int i=0;
	while(num>>1){
		i++;
		num=num>>1;
	}
	b[i]++;
	return ;
}//统计最高位为i的数目 
int main(){
	int cases,n,i,j,ans;
	scanf("%d",&cases);
	while(cases--){
		memset(b,0,sizeof(b));//初始化 
		ans=0;
		scanf("%d",&n);
		for(i=0;i<n;i++){
			scanf("%d",&a[i]);
			caculate(a[i]);
		}
		for(i=0;i<n;i++){
			j=31;//就是32位下int的最大值 即2的31次方
			while(j>=0){
				if(a[i]&(1<<j))break;//找到和a[i]最高位相同的数 
				j--;
			}
			while(j>=0){
				if(!(a[i]&(1<<j))){ans+=b[j];}
				j--;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}