D. Repetitions Decoding

// Problem: D. Repetitions Decoding
// Contest: Codeforces - Codeforces Round #773 (Div. 2)
// URL: https://codeforces.com/contest/1642/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-02-25 13:11:50
// 
// Powered by CP Editor (https://cpeditor.org)

#include
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
vector ins;
vectora;
vector lens;
void solve() {
	a.clear();
	ins.clear();
	lens.clear();
	int n;
	cin >> n;
	for (int i = 0; i < n ; i++) {
		int x;
		cin >> x;
		a.pb(x);
	}
	while (!a.empty()) {
		int i = a.size() - 1;
		int j = i - 1;
		while (j >= 0 && a[j] != a[i]) j --;
		if (j == - 1) {
			puts("-1");
			return;
		}
		for (int k = 0; k < i - j - 1; k ++) {
			ins.emplace_back(mp(j + k, a[i - 1 - k]));
		}
		lens.pb(2 * (i - j));
		reverse(a.begin() + j + 1, a.begin() + i);
		a.pop_back();
		a.erase(a.begin() + j);
	}
	reverse(all(lens));
	cout << ins.size() << endl;
	for (auto [p, c] : ins) {
		cout << p << " " << c << endl;
	}
	cout << lens.size() << endl;
	for (auto t: lens) cout << t<< " n";
	puts("");
}
int main () {
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

如何构造 选择两个相同的数，然后删除，接着把两个相同之间的数反转放到后一个数的后面。 这里代码是从大到小实现的 通过找数量规律，实现代码