D. Repetitions Decoding

D. Repetitions Decoding

// Problem: D. Repetitions Decoding// Contest: Codeforces - Codeforces Round #773 (Div. 2)// URL: Memory Limit: 256 MB// Time Limit: 1000 ms// 2022-02-25 13:11:50// // Powered by CP Editor (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_pairconst 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

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