CodeForces 1733C Parity Shuffle Sorting

题目要求单调不减，同时也不要求操作次数最少。一次操作可以改变一个数字，最少操作 $n$ 次可以对所有的数字都修改一次。如果所有数字相等也可以满足条件，所以利用这一点，先把 $a_1$ 和 $a_n$ 操作成相等的元素，然后对中间的每个数字 $a_i$，根据 $a_i + a_1$ 的奇偶性来让 $a_i$ 和 $a_1$ 或者 $a_n$ 配对，这样最后所有的元素都相等。

// Date: Sat Nov 25 20:41:10 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 100010;
int t, n, a[N];

int main(void) {
#ifdef _DEBUG
  freopen("1733c.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    For1(i, 1, n) { cin >> a[i]; }

    if (n == 1) {
      cout << "0\n";
      continue;
    }

    vector<PII> res;
    res.pb({1, n});
    if ((a[1] + a[n]) & 1)
      a[n] = a[1];
    else
      a[1] = a[n];

    For1(i, 2, n - 1) {
      int tmp = a[1] + a[i];
      if (tmp & 1)
        res.pb({1, i});
      else
        res.pb({i, n});
    }

    int m = SZ(res);
    cout << m << '\n';

    if (m) {
      For(i, 0, m) { cout << res[i].f1 << ' ' << res[i].f2 << '\n'; }
    }
  }

  return 0;
}