Submission #270528

#	Time	Username	Problem	Language	Result	Execution time	Memory
270528		s0m30n3	Binary Subsequences (info1cup17_binary)	C++14	100 / 100	718 ms	540 KiB

binary

#include<bits/stdc++.h>
#define sz(x) ((int)x.size())
#define pb push_back
#define ii pair<int,int>
#define iii pair<ii,int>
#define ppb pop_back
#define st first
#define nd second
#define ll long long
#define inf 1000000000
#define MOD 1000000007 
#define LOG 40
#define EPS 0.000000001
#define MAX 1000005
using namespace std;

int solve_reverse(int fix_0, int fix_1) {

	/*
		Our base case is dp[0][0] = dp[0][1] = 0
	*/
	int n = 0;
	while(max(fix_0, fix_1) > 0) {
		/*	The condition of while loop
			indicates whether we reached
			the base case mentioned above
		*/
		if(fix_0 == fix_1) {
			/*
				This line basically says 
				it is impossible to have such
				situation.
				Because both 
				fix_0 + 1 > fix_1 and
				fix_1 + 1 > fix_0.
				So this is a non-reachable
				state.
			*/
			return inf;
		}
		if(fix_0 > fix_1) {
			n += fix_0 / (fix_1 + 1);
			/*	This line speeds up the operations
				which are decreasing fix_1 + 1 from 
				fix_0 whenever the 'if' condition is
				satisfied.
			*/
			fix_0 %= fix_1 + 1;
			/*	this line gives us the result of the
				sped up process.
			*/
		}
		else {
			n += fix_1 / (fix_0 + 1);
			// similar
			fix_1 %= fix_0 + 1;	
			// similar
		}
	}
	// yeah, we found n !!!
	return n;

}

void solve() {

	int k, tot = 0, best = inf;
	int dp_fix_0 = -1;
	cin >> k;
	// fix dp[n][0]
	for(int dp_0 = 0; dp_0 <= k; dp_0++) {
		/*	automatically
			dp[n][1] = k - dp[n][0]
		*/
		int n = solve_reverse(dp_0, k - dp_0); 
		/*	now we have 'n' which is
			the length of the string
			if dp[n][0] was fixed like that
		*/
		if(n != inf) { 
			// if this dp[n][0] is valid
			tot++; 
			if(n < best) {
				// if this n is smallest
				best = n;
				dp_fix_0 = dp_0;
			}
		}
	}
	cout << tot << "\n";
	if(tot == 0) {
		cout << -1 << "\n";
		return ;
	}
	// let's simulate the solve_reverse function
	int dp_fix_1 = k - dp_fix_0;
	
	vector<char> result;
	
	while(max(dp_fix_0, dp_fix_1) > 0) {
		/*	this section is the straightforward 
			implementation of the 
			'solve_reverse' function.
		*/
		if(dp_fix_0 > dp_fix_1) {
			dp_fix_0 -= dp_fix_1 + 1;
			result.pb('0');
		}
		else {
			dp_fix_1 -= dp_fix_0 + 1;
			result.pb('1');
		}

	}
	reverse(result.begin(), result.end());
	// reverse the result because we reverse the actual process
	for(auto c: result) 
		cout << c << " ";
	cout << "\n";

}

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);

	int t;
	cin >> t;
	while(t--)
		solve();

}

• Subtask 1: 43 / 43
• Subtask 2: 39 / 39
• Subtask 3: 18 / 18