Submission #256273

#	Time	Username	Problem	Language	Result	Execution time	Memory
256273		mode149256	Type Printer (IOI08_printer)	C++14	100 / 100	76 ms	5864 KiB

printer

/*input
3
print
the
poem
*/
#include <bits/stdc++.h>
using namespace std;

namespace my_template {
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}
}
using namespace my_template;

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;

vector<char> ats;
string curr(MX, '-');
str did = "";

int j = 1;
void del() {
	assert(j > 0);
	ats.emplace_back('-');
	j--;
}

void add(char ch) {
	curr[j++] = ch;
	ats.emplace_back(ch);
}

void print() {
	ats.emplace_back('P');
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N;
	cin >> N;
	vs sk(N);
	for (int i = 0; i < N; ++i) {
		cin >> sk[i];
		sk[i].insert(sk[i].begin(), '-');
		if (sk[i].size() > did.size())
			did = sk[i];
	}

	sort(sk.begin(), sk.end(), [](const str & a, const str & b) {
		int k1 = 0;
		int k2 = 0;

		while (k1 + 1 < (int)did.size() and a[k1 + 1] == did[k1 + 1]) k1++;
		while (k2 + 1 < (int)did.size() and b[k2 + 1] == did[k2 + 1]) k2++;


		if (k1 != k2) return k1 < k2;
		else return a < b;
	});

	for (int i = 0; i < N; ++i)
	{
		int k = 0;
		str s = sk[i];

		while (k + 1 < (int)s.size() and s[k + 1] == curr[k + 1]) {
			k++;
		}

		while (j - 1 > k) {
			del();
		}

		// printf("k = %d, j = %d,ssz = %d\n", k, j, (int)s.size());
		for (int r = j; r < (int)s.size(); ++r)
		{
			add(s[r]);
		}
		print();
	}

	// printf("SUCCESS\n");
	printf("%d\n", (int)ats.size());
	for (int i = 0; i < (int)ats.size(); ++i)
	{
		printf("%c\n", ats[i]);
	}
}

/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 10 / 10