CHEFSIGN - Editorial

ad-hoc
cakewalk
chefsign
editorial
greedy
july17

#1

PROBLEM LINK:

Practice
Contest

Author: Dmytro Berezin
Primary Tester: Misha Chorniy
Editorialist: Hussain Kara Fallah

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

Given a sequence of N arithmetic signs ( = , + , - ) between N+1 unknown integers. You are allowed to fix these integers using numbers in the range [1,P] such that the expression is true when you read it from left to right. (Numbers don’t violate the signs). You are asked to calculate minimum possible P such you can construct a valid expression.

EXPLANATION:

First of all, it’s clear that we can drop all ‘=’ signs and pretend that they never existed, because each number which follows an ‘=’ sign would be identical to the number preceding this sign. So these signs aren’t affecting our answer at all.
After discarding all ‘=’ signs our answer would be :

P = max(maximum number of consecutive ‘<’ signs, maximum number of consecutive ‘>’ signs) + 1

Let’s process our expression from left to right, If we are facing X (X ≤ P-1) consecutive ‘<’ signs, our starting number would be P-X, and we increment our last number by 1 after each sign,so the number after the last sign would be exactly P (which will be followed by ‘>’ sign). Our last number will be followed by Y consecutive ‘>’ signs, so we assign the next number to Y (Y < P) and we decrement the last number by 1 by each ‘>’ sign we process. The number after the last sign would be 1. (In case our expression starts with ‘>’ the situation would just be reversed).
After that we would have another block of Z consecutive ‘<’ signs, so we assign the next number to P-Z (P-Z ≥ 1) so the number after the last sign would be P and we continue…

Following this approach, the last number after a block of consecutive ‘<’ signs would be P (the maximal value), and the last number after a block of consecutive ‘>’ signs would be 1 (the minimal value). So according to our bold assumption below we can assign values using numbers in the range [1,P] without violating the signs.

Let’s take an example:

<<<=>=>=>>><<>>>><<<<<>><<>>

After removing = signs our sequence would be

<<< >>>>> << >>>> << >>>>>

(Blocks are separated by spaces for clarity)

here P = max(3 , 5 , 2 , 4 , 5 , 2 , 2 , 2) + 1 = 6

Our sequence would be

3 < 4 < 5 < 6 > 5 > 4 > 3 > 2 > 1 < 5 < 6 > 4 > 3 > 2 > 1 < 5 < 6 > 5 > 4 > 3 > 2 > 1

AUTHOR’S AND TESTER’S SOLUTIONS:

TESTER’s solution: Will be found here
EDITORIALIST’s solution: Will be found here


#2

could any one tell why this gives WA https://www.codechef.com/viewsolution/14569995


#3

I am also getting few of the testcases accepted and others wrong answer.
https://www.codechef.com/viewsolution/14563406

Please help me find my mistake!

#include <iostream>
#define lli long long int
using namespace std;

int main(void) {
    int t;
    cin >> t;
    while(t--){
        char s[100011];
        lli tmp = 1, pmax = 1, pmin = 1;
        lli i;
        lli ans;
        cin >> s;
        for(i = 0; s* != '\0'; i++){
            if(s* == '<'){
                tmp++;
            } else if(s* == '>'){
                tmp--;
            }
            if(pmax < tmp) pmax = tmp;
            if(pmin > tmp) pmin = tmp;
        }
        //cout << pmax << " " << pmin << endl;
        if(pmin < 1) {
            ans = pmax - pmin + 1;
        } else {
            ans = pmax;
        }
        cout << ans << endl;
    }
    return 0;
}

#4

Yes, In both of your codes, you guys are assuming that the numbers increase or decrease by 1, which is not always the case.
Consider this,
str = ‘><<<><<’

Your code will give the answer as 5.
your answer == 2>1<2<3<4>3<4<5

but the actual answer would be 4
== 2>1<2<3<4>1<2<3

like this.


#5

Here is my code … what do you think is wrong about it. Please, help me find my mistakes

#include <bits/stdc++.h>

using namespace std;

#define fast_cin ios_base::sync_with_stdio(false)

typedef long long ll;

int main()
{
	fast_cin;
	ll t; cin >> t;
	while(t--){
		string s; cin >> s;
		ll in = 1, ans = 0;
		for (int i = 0; i < s.size(); ++i)
		{
			if (s* == '<')
			{
				in++;
				ans = max(ans,in);
			}
			else if (s* == '>')
			{
				in = 1;
			}
		}
		cout << ans << endl;
	}

	return 0;
}

It is giving current answer for string s = '<<<=>=>=>>><<>>>><<<<<>><<>>'or ‘><<<><<’


#6

can any one pls tell me why am i not able to pass 2 cases in main task… ? i have my code and screen shot of my result here…

https://pastebin.com/6QJRau0v


#7

Can anybody tell me what’s wrong with it.

https://www.codechef.com/viewsolution/14523196