Codeforces Round #555 (Div. 3) Editorial

#	User	Rating
1	tourist	3557
2	Radewoosh	3468
3	Um_nik	3429
4	Petr	3354
5	Benq	3286
6	mnbvmar	3280
7	wxhtxdy	3276
7	LHiC	3276
9	ecnerwala	3214
10	yutaka1999	3190

#	User	Contrib.
1	Errichto	191
2	Radewoosh	180
3	tourist	173
4	Vovuh	165
4	antontrygubO_o	165
6	PikMike	164
7	rng_58	160
8	Um_nik	156
8	majk	156
10	300iq	154

Tutorial

Solution

def f(x):
	x += 1
	while(x % 10 == 0):
		x //= 10
	return x

a = set()
n = int(input())

while(not(n in a)):
	a.add(n)
	n = f(n)

print(len(a))

1157B - Long Number

Idea: BledDest

Tutorial

Solution

#include<bits/stdc++.h>

using namespace std;

int f[10];
string s;

int main()
{
	int n;
	cin >> n;
	cin >> s;
	for(int i = 1; i <= 9; i++)
		cin >> f[i];
	vector<int> diff;
	for(int i = 0; i < n; i++)
		diff.push_back(f[s[i] - '0'] - (s[i] - '0'));
	for(int i = 0; i < n; i++)
		if(diff[i] > 0)
		{
			while(i < n && diff[i] >= 0)
			{
				s[i] = char(f[s[i] - '0'] + '0');
				i++;
			}
			break;
		}
	cout << s << endl;
}

1157C1 - Increasing Subsequence (easy version)

Idea: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	
	string res;
	int l = 0, r = n - 1;
	int lst = 0;
	while (l <= r) {
		vector<pair<int, char>> cur;
		if (lst < a[l]) cur.push_back(make_pair(a[l], 'L'));
		if (lst < a[r]) cur.push_back(make_pair(a[r], 'R'));
		sort(cur.begin(), cur.end());
		if (int(cur.size()) == 2) {
			cur.pop_back();
		}
		if (int(cur.size()) == 1) {
			if (cur[0].second == 'L') {
				res += 'L';
				lst = a[l];
				++l;
			} else {
				res += 'R';
				lst = a[r];
				--r;
			}
		} else {
			break;
		}
	}
	
	cout << res.size() << endl << res << endl;
	
	return 0;
}

1157C2 - Increasing Subsequence (hard version)

Idea: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	
	string res;
	int l = 0, r = n - 1;
	int lst = 0;
	while (l <= r) {
		vector<pair<int, char>> cur;
		if (lst < a[l]) cur.push_back(make_pair(a[l], 'L'));
		if (lst < a[r]) cur.push_back(make_pair(a[r], 'R'));
		sort(cur.begin(), cur.end());
		if (int(cur.size()) == 2 && cur[0].first != cur[1].first) {
			cur.pop_back();
		}
		if (int(cur.size()) == 1) {
			if (cur[0].second == 'L') {
				res += 'L';
				lst = a[l];
				++l;
			} else {
				res += 'R';
				lst = a[r];
				--r;
			}
		} else if (int(cur.size()) == 2) {
			int cl = 1, cr = 1;
			while (l + cl <= r && a[l + cl] > a[l + cl - 1]) ++cl;
			while (r - cr >= l && a[r - cr] > a[r - cr + 1]) ++cr;
			if (cl > cr) {
				res += string(cl, 'L');
			} else {
				res += string(cr, 'R');
			}
			break;
		} else {
			break;
		}
	}
	
	cout << res.size() << endl << res << endl;
	
	return 0;
}

1157D - N Problems During K Days

Idea: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n, k;
	cin >> n >> k;
	if (n < k * 1ll * (k + 1) / 2) {
		cout << "NO" << endl;
		return 0;
	}
	
	int nn = n - k * (k + 1) / 2;
	vector<int> a(k);
	for (int i = 0; i < k; ++i) {
		a[i] = i + 1 + (nn / k) + (i >= k - nn % k);
	}
	
	if (nn != k - 1) {
		cout << "YES" << endl;
		for (int i = 0; i < k; ++i) cout << a[i] << " ";
		cout << endl;
	} else {
		if (k > 3) {
			--a[1];
			++a[k - 1];
		}
		if (k == 2 || k == 3) {
			cout << "NO" << endl;
		} else {
			cout << "YES" << endl;
			for (int i = 0; i < k; ++i) cout << a[i] << " ";
			cout << endl;
		}
	}
	
	return 0;
}

1157E - Minimum Array

Idea: Vovuh

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n;
	cin >> n;
	vector<int> a(n);
	multiset<int> b;
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	for (int i = 0; i < n; ++i) {
		int x;
		cin >> x;
		b.insert(x);
	}
	
	for (int i = 0; i < n; ++i) {
		auto it = b.lower_bound(n - a[i]);
		if (it == b.end()) it = b.begin();
		cout << (a[i] + *it) % n << " ";
		b.erase(it);
	}
	cout << endl;
	
	return 0;
}

1157F - Maximum Balanced Circle

Idea: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n;
	cin >> n;
	vector<int> a(n);
	vector<int> cnt(200 * 1000 + 1);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
		++cnt[a[i]];
	}
	
	sort(a.begin(), a.end());
	a.resize(unique(a.begin(), a.end()) - a.begin());
	
	int l = 0, r = 0;
	int ans = cnt[a[0]];
	for (int i = 0; i < int(a.size()); ++i) {
		int j = i + 1;
		int sum = cnt[a[i]];
		while (a[j] - a[j - 1] == 1 && cnt[a[j]] > 1) {
			sum += cnt[a[j]];
			++j;
		}
		int cr = j - 1;
		if (j < n && a[j] - a[j - 1] == 1) {
			sum += cnt[a[j]];
			cr = j;
		}
		if (ans < sum) {
			ans = sum;
			l = i;
			r = cr;
		}
		i = j - 1;
	}
	
	cout << ans << endl;
	for (int c = 0; c < cnt[a[l]]; ++c) cout << a[l] << " ";
	for (int i = l + 1; i < r; ++i) {
		for (int c = 0; c < cnt[a[i]] - 1; ++c) cout << a[i] << " ";
	}
	for (int c = 0; l != r && c < cnt[a[r]]; ++c) cout << a[r] << " ";
	for (int i = r - 1; i > l; --i) cout << a[i] << " ";
	cout << endl;
	
	return 0;
}

1157G - Inverse of Rows and Columns

Idea: Vovuh

This is the comment about the quadratic solution. Thank you so much for mentioning this fact, STommydx!

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int n, m;

void invRow(vector<vector<int>> &a, int idx) {
	for (int pos = 0; pos < m; ++pos) {
		a[idx][pos] ^= 1;
	}
}

void invCol(vector<vector<int>> &a, int idx) {
	for (int pos = 0; pos < n; ++pos) {
		a[pos][idx] ^= 1;
	}
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	cin >> n >> m;
	vector<vector<int>> a(n, vector<int>(m));
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < m; ++j) {
			cin >> a[i][j];
		}
	}
	
	for (int cnt0 = 0; cnt0 <= m; ++cnt0) {
		string r(n, '0');
		string c(m, '0');
		vector<vector<int>> b = a;
		for (int j = 0; j < m; ++j) {
			if ((j < cnt0 && b[0][j] == 1) || (j >= cnt0 && b[0][j] == 0)) {
				invCol(b, j);
				c[j] = '1';
			}
		}
		bool ok = true;
		if (cnt0 < m) {
			for (int i = 1; i < n; ++i) {
				int sum = accumulate(b[i].begin(), b[i].end(), 0);
				if (sum != 0 && sum != m) {
					ok = false;
				} else if (sum == 0) {
					invRow(b, i);
					r[i] = '1';
				}
			}
		} else {
			int idx = -1;
			for (int i = 1; i < n; ++i) {
				int sum = accumulate(b[i].begin(), b[i].end(), 0);
				if (sum != 0 && sum != m) {
					if (idx != -1) ok = false;
					else idx = i;
				}
			}
			if (idx == -1) {
				for (int i = 1; i < n; ++i) {
					int sum = accumulate(b[i].begin(), b[i].end(), 0);
					if (sum == 0) {
						invRow(b, i);
						r[i] = '1';
					}
				}
			} else {
				for (int i = 1; i < idx; ++i) {
					int sum = accumulate(b[i].begin(), b[i].end(), 0);
					if (sum == m) {
						invRow(b, i);
						r[i] = '1';
					}
				}
				if (b[idx][0] == 1) {
					invRow(b, idx);
					r[idx] = '1';
				}
				for (int j = 1; j < m; ++j) {
					if (b[idx][j] < b[idx][j - 1]) {
						ok = false;
					}
				}
				for (int i = idx + 1; i < n; ++i) {
					int sum = accumulate(b[i].begin(), b[i].end(), 0);
					if (sum == 0) {
						invRow(b, i);
						r[i] = '1';
					}
				}
			}
		}
		if (ok) {
			cout << "YES" << endl << r << endl << c << endl;
			return 0;
		}
	}
	
	cout << "NO" << endl;
	
	return 0;
}

include<bits/stdc++.h>

using namespace std; int main() { char x[10]; int n,i,f=0,t1,t2; string p; cin >> n; cin >> p; for(i=1;i<=9;i++) cin >> x[i]; for(i=0;i<p.size();i++) { t1=p[i]-'0'; t2=x[t1]-'0'; // cout << t1 << " " << t2 << endl; if(t2>t1) { p[i]=x[t1]; f=1; } else if(f==1) { break; }

} cout << p << endl;

}

Comments (37)

Write comment?

DGC

6 months ago, # |

The editorial for Problem G looks like a little the editorial of Problem F

→ Reply

Vovuh

6 months ago, # ^ |

Thank you for mentioning, I just forgot to fix this place :)

It_Wasnt_Me

it was a greedy contest :D Good round by the way

sodafago

My solution problem D after contest(sad): 1)Find minimum first element with binary search 2) let's get maximum suffix, when we can add 1 for all numbers on segment, do it, while exist this is segment 3) end O(nLogn), and write easy

https://codeforces.com/contest/1157/submission/53393728

F.E.A.R

← Rev. 5 →

In problem "A" we don't need to store reachable numbers ,As soon as we hit a number less than 10 we add 9 to our counter and stop

Nice problems btw

myHandle

1157D — N Problems During K Days can be solved simply by putting $$$1,2,\dots,k$$$ initially and then trying to put $$$(n-(k*(k+1)/2))/k$$$ elements in each, and then from reverse order adding as many elements in each position as possible, without considering special cases separately.

You can go through my submission for clear understanding.

Mayank612512

In problem number B, why we are not checking the digits which comes after the first digit which become less after replacing.

for example 
if a = 13373
and f[] = {1, 2, 5, 4, 6, 6, 3, 1, 9};
then the correct answer is 15573 but why not 15575;

geekotechy

The problem mentions a contiguous subsegment and the operation can be performed at most once.

AmericanViper9022

Anyone help me in finding the cause of TLE in Problem D in my solution.

sorry_marymarine

← Rev. 2 →

"erase" for vector works in linear time of size of the whole vector

And what about erase in multiset?

Is there any way that I can still use vector and solve the problem. I know that using multiset will make it very easy but still as a matter of knowing.

BTW, Thanks for the reply buddy !!

Aluss

You can’t use vector.erase for this problem. unless you want to code yourself a multiset libary for vectors, multiset is the only way to solve the problem in AC time, which erases elements from the container at log(n) time.

Thanx for the reply mate !!

vjvjain0

This problem also has a solution like D:

https://www.codechef.com/SNCK1B19/problems/MAXPRODU

Michelin_star

please help me...y its showing error in test case 15 always!!!

https://codeforces.com/contest/1157/submission/53440011

aisekaise

make it a habit to write long codes, your is too short to get accepted

prac64

Does anyone have a solution for problem E which does not involve STL structures, specifically without using multiset,set,map or multimap ?

islammohsen

You can use segment trees.

check this solution 53473452

ankyy_0103

Help me out in finding TLE in my Solution for Problem C

mjd-dalloul

while(l <= r)

Vishal_verma

In problem E, why "lower_bound(m.begin(), m.end(), val)" gives TLE 53488500 and "m.lower_bound(val)" got Accepted 53488794 ?

got it! https://codeforces.com/blog/entry/20553

Naman_Bhalla

Problem D: "Then if nn != k — 1 or k = 1 then this answer is correct."

Any explanation for this ?

Kangmin_Tan

For problem 1157c1, what if the input is

7 7 4 5 6 3 2 1

The solution code out puts 5 RRRRL But if I discard 7 and take the subsequence [4 5 6 3 2 1],it's possible to get a strictly increasing subsequence of size 6, right?

fengquyoumo

When is the next div3?

Mohit_Thorat

What's wrong with my solution for problem B? https://codeforces.com/contest/1157/submission/53566680 I am getting wrong answer on test case 7

buzz_95

Try for
5
43111
9 2 3 5 1 1 1 1 1
Answer should be : 53999, get it?

GaryMr

I solved D in a different way. Mine

vlado

I'm kind of late, but problem D can be easily (more or less) solved by binary searching the first element, and then binary searching every other element as well, from left to right, so that the minimum possible sum is <= n and the maximum possible sum is >= n. My submission

InshaAllah

Nice Problem Set.

For the D problem, I just initialized array to 1,2,...,K then added (S — (K * (K + 1)) / 2) / N to every element in the array, and then just distributed (S — (K * (K + 1)) / 2) % N from the behind appropriately. Just wondering why there were so less submissions for this problem?

codershuvo

How to solve problem Minimum Array using segment trees??

Rockett

I solved with segment tree, so I will try to explain my solution:

First we need to sort B array, to be able to find some index j for each $$$A_i$$$ such that $$$A_i + B_l \lt n$$$ for all $$$l \lt j$$$, and $$$A_i + B_r \ge n$$$ for all $$$r \ge j$$$. We'll have only one index j for each $$$A_i$$$, because $$$max(A) + max(B) < 2n$$$.

After that we can divide B array in two parts and find the best among these greedly, using segment tree here to get minimum, and then updating the used element to not be taken more than once.

Submission

harshraj22

5 months ago, # |

can anyone please tell why my recursive solution for c2 gives tle ? I think my solution too has the same time complexity as the editorial's .

kuskmen

Guys can someone help me understand why http://codeforces.com/contest/1157/submission/53985650 gives wrong results on test case 9

Maskur_ICT_MBSTU

2 weeks ago, # |

Your code here...

//why it doesn't work for problem B

/*what's going on the problem C2 it give the right answer for test case 1 in online compiler but in the cf compiler shows wrong answer.I submitted this code in gnu c++17 */ https://codeforces.com/contest/1157/submission/61860615

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include<bits/stdc++.h>