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n = float(raw_input())
lst = [int(x) for x in raw_input().split()]
print format(len([x for x in lst if x > 0])/n, ".6f")
print format(len([x for x in lst if x < 0])/n, ".6f")
print format(len([x for x in lst if x == 0])/n, ".6f")
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
double a=0,b=0,c=0;
int arr[] = new int[n];
for(int arr_i=0; arr_i < n; arr_i++){
arr[arr_i] = in.nextInt();
}
for(int arr_i=0; arr_i < n; arr_i++){
if (arr[arr_i]>0)
{a++;}
else if (arr[arr_i]<0){b++;}
else if (arr[arr_i]==0){c++;}
}
System.out.println(a/n);
System.out.println(b/n);
System.out.println(c/n);
}
Could have done the conditionals and assigning arr the scanner.nextInt() numbers all in one for loop:
importjava.io.*;importjava.util.*;publicclassSolution{publicstaticvoidmain(String[]args){/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */Scannerscanner=newScanner(System.in);intsize=scanner.nextInt();int[]a=newint[size];doublefractionPositive=0;doublefractionZero=0;doublefractionNegative=0;for(inti=0;i<size;i++){a[i]=scanner.nextInt();if(a[i]>0)fractionPositive+=1;elseif(a[i]==0)fractionZero+=1;elseif(a[i]<0)fractionNegative+=1;}System.out.println(fractionPositive/size);System.out.println(fractionNegative/size);System.out.println(fractionZero/size);}}
You also don't need to count one of them, for instance, if you don't count negative ones, your output is:
System.out.println((size-fractionPositive-fractionZero)/size);
Most programing languages do not like when variable types mix with eachother. An integer divided by an integer produces a floating point in most cases, but the computer only works with variables that all have the same data type; such as a double divided by a double producing a double. The problem seems like you are having is because of inconsistant varible types.
The folowing case does not work because there are two differant variable types:
double = integer / integer
A solution to this problem is to use a STATIC CAST. The following line could be read as "turn my (positiveIntegerCount) into a double before i divide by its length":
Plus Minus
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Quick help? How do I format code snippet in my reply? I used
and it isn't formatted. Thanks in advance.Nevermind! got it from a comment below :)
n = float(raw_input()) lst = [int(x) for x in raw_input().split()] print format(len([x for x in lst if x > 0])/n, ".6f") print format(len([x for x in lst if x < 0])/n, ".6f") print format(len([x for x in lst if x == 0])/n, ".6f")
public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); double a=0,b=0,c=0; int arr[] = new int[n]; for(int arr_i=0; arr_i < n; arr_i++){ arr[arr_i] = in.nextInt(); }
Could have done the conditionals and assigning arr the scanner.nextInt() numbers all in one for loop:
Should've used printf for more precise result. Cool solution tho'.
https://www.hackerrank.com/challenges/plus-minus/forum/comments/236802
is it ("%.6f % n") regular expression and how it is formatting the result
Ahh got it to limit the output after decimal..
but in between " " this will coment than how wil be this work??
thank u...
("%.6f %n", this "," sould we use +
no this will print the whole string
import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
} why this code is not being approved???
because there is one \n at the beginning which is not necasary
System.out.printf("%1$.6f", p/al);
add \n after 6f in each s.o.pln then it will pass all test cases
just remove \n from first printf not add !
just remove \n from first printf as its starting from end of line which is against the format
cool solution bro
We can also use float instead of double. It will work.
Didn't really need an array.
You also don't need to count one of them, for instance, if you don't count negative ones, your output is: System.out.println((size-fractionPositive-fractionZero)/size);
This could be used as a replacement for your print statement
good solution to union all sout.
Can someone exaplin how to get this regex? %.06f%n%.06f%n%.06f I am a little confused
It is not a regex, Google pls - print format in Java
awesome thanks
where did you set the precision in this code??
i am trying to do the same thing but the output remains 0.000000 i dont know how?
Are your variables of type
doubleinstead ofint?no.
Same for me...
yea.. same for me also...
code is working in VS IDE, but submited code here not passing the test cases. Don`t why. Submitted code is correct
did you round the results of the division?
Most programing languages do not like when variable types mix with eachother. An integer divided by an integer produces a floating point in most cases, but the computer only works with variables that all have the same data type; such as a double divided by a double producing a double. The problem seems like you are having is because of inconsistant varible types.
The folowing case does not work because there are two differant variable types: double = integer / integer
A solution to this problem is to use a STATIC CAST. The following line could be read as "turn my (positiveIntegerCount) into a double before i divide by its length":
typecast into float
same here....!
apply casting while dividing number of positive(negative or zeroes)by number of elements e.g. a=(float)b/n;
python 2 maybe?
This won't give decimal output
but what is data type of a&b&c???