**Problem:-**Plus Minus solution in C++ or Plus Minus hacker rank solution in Java or Plus Minus hacker rank solution c++ or plus-minus hacker rank solution c++ or Plus Minus program in c or Plus Minus solution in c or Plus Minus hacker rank solution in python or Plus-Minus in java or Plus Minus hacker rank solution in c or Hacker Rank Solution For Plus-Minus in C++ or Plus-Minus in C++ or Hacker Rank solution for Plus Minus in C++.

**Logic:-**Plus Minus is a very simple problem we have to just count positive number, Negative number, and Number that is equal to zero, so we can do this by using an array or vector and put the three condition if the number is greater than zero increase a count and if the number is less than zero increase second count variable and if both conditions are not true that increase the third count variable.

At the end of the solution divide all three count variables by a total number of a variable taken from input time in an array or vector. See the explanation with an example for a better understanding.

Now put the second condition if the number is less than zero then increase a count2 variable by 1 each time.

And if both the conditions are not true then increase the third count or count3 variable by 1 each time.

Now we have all three count values or we also know the total number of values in an array, so just print the below statement. Here

print c1 / n

print c2 / n

print c3 / n

Hence we did it we got a solution.

**Explanation:-**First put the condition if the number is greater than zero then increase the count1 variable by 1 each time.**if(arr[arr_i]>0)****{****c1++;****}**Now put the second condition if the number is less than zero then increase a count2 variable by 1 each time.

**if(arr[arr_i]<0)****{****c2++;****}**And if both the conditions are not true then increase the third count or count3 variable by 1 each time.

**if(arr[arr_i]==0)****{****c3++;****}**Now we have all three count values or we also know the total number of values in an array, so just print the below statement. Here

**n**is a total number of an element in an array. Check C++ Homework Help.print c1 / n

print c2 / n

print c3 / n

Hence we did it we got a solution.

**Submit your solution here:-**Click here

**Tip:-**Add

**setprecision(6)**before printing the output on-screen cause we have to print our solution up to 6 decimal digits. Before copying the solution I recommended please you read this full article, this will help you to build your own logic.

**Solution:-**

#include <bits/stdc++.h>

using namespace std;

int main()

{

int n,c1=0,c2=0,c3=0;

cin >> n;

vector<int> arr(n);

for(int arr_i = 0;arr_i < n;arr_i++)

{

cin >> arr[arr_i];

if(arr[arr_i]>0)

{

c1++;

}

if(arr[arr_i]<0)

{

c2++;

}

if(arr[arr_i]==0)

{

c3++;

}

cout<<setprecision(6)<<(float)c1/n<<endl;

cout<<setprecision(6)<<(float)c2/n<<endl;

cout<<setprecision(6)<<(float)c3/n<<endl;

return 0;

}

**Output:-**

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