Temperatures puzzle discussion


#694

(PHP) Tous les tests passent dans la page de code à 100%. et seulement 85 % après soumission sur question 2 et 3.
(PHP) all code example passed in dev hud. but after submiting only 85 % . any clue ?


#695
 int CT=abs(temp[0]);
    int j=sizeof(temp);
    for(int i=0;i<n;i++)
    {
        if(CT>abs(temp[i]))
        {
            CT=abs(temp[i]);
             p=i;
            
        
        }
    }
    for(int i=0;i<n;i++)
    {
        if(CT==abs(temp[i]))
        {
            if(temp[p]<temp[i])
            {
                CT=temp[i];
            }
        }
        else
        {
            CT=temp[p];
        }
    }

whats the problem with this code i cant get through the the 3 test case everything else does work


#696

I advise you to add “error prints” to your algorithm to know the values of CT and p for example. You’ll quickly see the mistake in your code. Good luck!


#697

Thanks! That did it. My code has a section that tests if the input is numeric, and sets the output to zero if that’s the case. The initial input values are string values, not integers. I ran intval() on the input value first, and it works now.


#698

I am stuck… I don’t know where is the bug.Can anyone help me out?

/**
* Auto-generated code below aims at helping you parse
* the standard input according to the problem statement.
**/

const n = parseInt(readline()); // the number of temperatures to analyse
var inputs = readline().split(' ');
var arr=[];
for (let i = 0; i < n; i++) {
    var t = parseInt(inputs[i]); // a temperature expressed as an integer ranging from -273 to 5526
     if(t<0){return 1}
    arr.push(t);
    
}

var closeToZero=Math.min(...arr);
// if(closeToZero<0){return 1}

// Write an action using print()
// To debug: printErr('Debug messages...');

print('closeToZero');

#699

You have to print an integer value, not a string.


#700

Please help me.
I dont understant the input of this programm.

We have n = nmbrs of temperature : n = int(raw_input()) ok
And we must have a string with the n temperatures, where is it ?
do we have to write it ?

Also I dont understant that line
for i in raw_input().split():
# t: a temperature expressed as an integer ranging from -273 to 5526
t = int(i)

because in this case t will be 1, 2 ,3, 4 , …

Why is it to be a string ? will not be better a list ?
So after we can take the min, of abs of the values in this list.

Thanks for your help


#701

Hey there I need some help in this,

I will revisit this later in the day that i am writing this but some tips or subtle hints would be greatly appreciated. This is logic that i have in the standard loop for the problem.

        if(i >= 1){
            if(Math.abs(pastNum) < Math.abs(t)){
                result = String.valueOf(pastNum);
            }
            
            else if (Math.abs(t) < Math.abs(pastNum)) {
                result = String.valueOf(t);
            }
            
            else if (Math.abs(pastNum) == Math.abs(t)){
                if(pastNum > 0){
                    result = String.valueOf(pastNum);
                }
                else if (t > 0){
                    result = String.valueOf(t);
                }
                
                else{
                    result = String.valueOf(t);
                }
            }
        }
        
        else{
            pastNum = t;
        }

Thanks again.


#702

So guys, I’m a total noob and trying to solve this… I’ve managed to get through most of the tests with the code I wrote (though it’s probably not the most convenient way) But I don’t see how I can pick the positive temperature instead of the negative one, any help on this?

my code:

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

/**
 * Auto-generated code below aims at helping you parse
 * the standard input according to the problem statement.
 **/
int main()
{
    int zero = 0;
    int result = 0 ;  
    int n; // the number of temperatures to analyse

    cin >> n; cin.ignore();
    
    //loop temperatures on row
    for (int i = 0; i < n; i++) {
        int t; // a temperature expressed as an integer ranging from -273 to 5526
        cin >> t; cin.ignore();
    
    if(i == 0 || (t < zero && ((zero - result) > (zero - t))) || ((t > zero && (zero + result) > (zero + t)))) {         
        result = t;}
        

    
    

    }

    // Write an action using cout. DON'T FORGET THE "<< endl"
    // To debug: cerr << "Debug messages..." << endl;

    cout << result << endl;

#703

tip: (https://www.codingame.com/training/easy/temperatures)

image


#704

Many thanks, did not know about the sqrt() function. Had to include the <math.h> library though. I’ve changed my if statement to the following and now it’s working!

if(sqrt(t*t) == sqrt(result*result)){
    result = sqrt(result*result);  
}
else if(i == 0 || (t < zero && ((zero - result) > (zero - t))) || ((t > zero && (zero + result) > (zero + t)))) {         
    result = t;}

#705

If you solved it to 100%, you have now access to published solutions (https://www.codingame.com/training/easy/temperatures/solution). I’m sure you will find more tricks to learn.


#706

Thanks, did not know about that yet. I now managed to get the abs() function in, instead of calculating the absolute integer myself. However, many of the functions I see in the solutions I haven’t familiarized myself yet, so I guess I’ll have to learn a little more before I can fully understand them, thanks again!


#707

Hi, I am using python3.

I haven’t been able to do any of the test case with any of the codes I have written. It always returns the furthest number. This is what I have.

import sys
import math

# Auto-generated code below aims at helping you parse
# the standard input according to the problem statement.

n = int(input())  # the number of temperatures to analyse

# print(type(input())) # String

if n == 0:
    closest=0
else:
    for i in (input().split()):
        # t: a temperature expressed as an integer ranging from -273 to 5526
        t = int(i)
        closest = 5527
        if abs(t)<int(closest):
            closest = t

# Write an action using print
# print("Debug messages...", file=sys.stderr)

print(closest)

#708

You can initialize closest with float(“inf”).
Moreover, “-5 5” will fail but “5 -5” will succeed.


#709

My code works well on any other website, but it’s not working on CodinGame. It works for 1,2,4, but it tells me it returns “12” for #3 and “-4” for #5. Any thoughts?

My code:

    n = int(input())  # the number of temperatures to analyse

    templist = {}
    realtemp = []
    abstemp = []
    newlist =[]


    for i in input().split():
        # t: a temperature expressed as an integer ranging from -273 to 5526
        a = int(i) # log the 
        t = (abs(int(i)))
        realtemp.append(a)
        abstemp.append(t)

    def dedupe(list):
        for m in list:
            if (list.count(m)+list.count(-m))>1:
                list[m]=abs(m)
        return list

    newtemp = dedupe(realtemp)

    templist= dict(zip(newtemp,abstemp))
    answer = min(templist, key=templist.get)

#710

it seems you just need to implement the following rule to validate all test cases:

If two numbers are equally close to zero, positive integer has to be considered closest to zero (for instance, if the temperatures are -5 and 5, then display 5).


#712

does any one know why (b) keep sending 0 as output ?
where and how to declare b to prevent this from happening?

#include <stdlib.h>
#include <stdio.h>
#include <string.h>


/**
 * Auto-generated code below aims at helping you parse
 * the standard input according to the problem statement.
 **/
int main()
{
    int n; // the number of temperatures to analyse
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        int t; 
         // a temperature expressed as an integer ranging from -273 to 5526
        scanf("%d",&t);
        printf("t=%d\n",t);// just for debugging
        if (abs(t)<b){b=t;}
        printf("b=%d\n",b);
        if (n==i) {printf("%d\n",b);}
    }

    // Write an action using printf(). DON'T FORGET THE TRAILING \n
    // To debug: fprintf(stderr, "Debug messages...\n");

    //printf("%d\n",b);

    return 0;
}

the output

### Standard Output Stream:

t=1

b=0

t=-2

b=0

t=-8

b=0

t=4

b=0

t=5

b=0

### Failure

Found:

t=...

Expected:

1

#714

B equals 0, everything is greater than it. you should initialise b to 9999 at the same place you initialise n.


#715

There is an error in the description or in the testcases in one of the games. You should accept both values that are same distance from the 0. There is no clarification that in case of two values being same distant from 0 we choose positive one. Not a valid testcase or not valid description.