c++写出下面代码 You are given a permutation a of length n. Find any permutation b of length n such that a1+b1≤a2+b2≤a3+b3≤…≤an+bn. It can be proven that a permutation b that satisfies the condition above always exists. †A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array), and [1,3,4] is also not a permutation (n=3 but there is 4 in the array). Input Each test contains multiple test cases. The first line of input contains a single integer t(1≤t≤2000) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer n (1≤n≤100) — the length of permutations a and b. The second line of each test case contains n distinct integers a1,a2,…,an(1≤ai≤n ) — the elements of permutation a. All elements of a are distinct. Note that there is no bound on the sum of n over all test cases. Output For each test case, output any permutation b which satisfies the constraints mentioned in the statement. It can be proven that a permutation b that satisfies the condition above always exists. Example input 5 5 1 2 4 5 3 2 1 2 1 1 3 3 2 1 4 1 4 3 2 output 1 2 4 3 5 2 1 1 1 2 3 1 2 3 4 Note In the first test case a=[1,2,4,5,3] . Then the permutation b=[1,2,4,3,5] satisfies the condition because 1+1≤2+2≤4+4≤5+3≤3+5.

以下是你要的代码：

#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN = 105;

int a[MAXN], b[MAXN];

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        sort(a, a + n);
        int l = 0, r = n - 1;
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) {
                b[i] = a[l];
                l++;
            } else {
                b[i] = a[r];
                r--;
            }
        }
        for (int i = 0; i < n; i++) {
            cout << b[i] << " ";
        }
        cout << endl;
    }
    return 0;
}

思路：首先将原排列 a 排序，然后将排序后的元素分成两部分，分别从前往后和从后往前取，这样可以保证 b 满足题目要求。