ZOJ3870 Team Formation【位运算+数学】

Team Formation

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Time Limit: 2 Seconds      Memory Limit: 131072 KB

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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. Thei^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

231 2 351 2 3 4 5

Sample Output

16

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Author: LIN, Xi
Source: The 12th Zhejiang Provincial Collegiate Programming Contest

问题链接：ZOJ3870 Team Formation。

问题简述：输入n个正整数，求任意两个数异或运算结果大于两个数中的最大值的组数。

问题分析：

这里给出异或运算的规则，1^1=0，1^0=1，0^1=1，0^0=0。

对于y而言，若y的第i位为1（从右到左分别为0-31位），y的第j位为0（j<i），

对于第j位为1并且是最高位的1的x，

那么y^x>y（按照假定满足x<y)。

程序说明：

数组bit[]，bit[i]表示各个数（所有的a）中第i位是最高位1的数的个数。

数组left[]，left[i]保存a[i]的为1的最高位。

题记：（略）

AC的C++语言程序如下：

/* ZOJ3870 Team Formation */#include <iostream>#include <stdio.h>#include <string.h>const int N = 100000;const int N2 = 32;int a[N], left[N], bit[N2];void left1count(int x, int i) {    int l = N2 - 1;    left[i] = -1;    while(l >= 0) {        if(x & (1 << l)) {            bit[l]++;            left[i] = l;            break;        }        l--;    }    return;}int main(){    int t, n;    scanf("%d", &t);    while(t--) {        memset(bit, 0, sizeof(bit));        scanf("%d",&n);        for(int i=0; i<n; i++) {            scanf("%d", &a[i]);            left1count(a[i], i);        }        long long ans = 0;        for(int i=0; i<n; i++) {            if(a[i]) {                while(left[i] >= 0) {                    if(!(a[i] & (1 << left[i])))                        ans += bit[left[i]];                    left[i]--;                }            }        }        printf("%lld\n", ans);    }    return 0;}