PROBLEM LINK:
Author: Setter’s name
Tester: Tester’s name
DIFFICULTY:
CAKEWALK,EASY.
PREREQUISITES:
Simple Array and Loops.
PROBLEM:
Baby Chef has an array A of n numbers. Output the array B where B[i] is the average of first i numbers in the array A.
QUICK EXPLANATION:
You can simply apply set an iterator i and calculate average of the array till that iterator and store it at position B[i].
EXPLANATION:
- First we will set an iterator that will iterate throughout the length of the array.
- Maintain a variable named sum.
- Iterating the array update the value of sum by operation sum = sum+ a[i].
- Finally store average in array B by B[i] = sum/(i+1).
SOLUTIONS:
Setter's Solution
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
vector<int>v;
int a;
vector<int>v1;
for(int i=0;i<n;i++)
{
cin>>a;
v.push_back(a);
}
int i = 1;
while(i<=n)
{
int avg=0;
int sum=0;
for(int x=0;x<i;x++)
sum=sum+v[x];
avg=(sum)/i;
v1.emplace_back(avg);
i++;
}
for(int i:v1)
cout<<i<<" ";
cout<<"\n";
}
return 0;
}
Tester's Solution
t = int(input())
for x in range(0,t):
size = int(input())
arr = map(int,input().split())
c = 1
sumn =0
for x in arr:
sumn +=x
print(sumn//c,end=" ")
c=c+1
print()