problem_solving_cpp/785A.cpp at master · Twinklebhatia/problem_solving_cpp · GitHub

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Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:

Tetrahedron. Tetrahedron has 4 triangular faces.
Cube. Cube has 6 square faces.
Octahedron. Octahedron has 8 triangular faces.
Dodecahedron. Dodecahedron has 12 pentagonal faces.
Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:


Anton has a collection of n polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!

#Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of polyhedrons in Anton's collection.

Each of the following n lines of the input contains a string si — the name of the i-th polyhedron in Anton's collection. The string can look like this:

"Tetrahedron" (without quotes), if the i-th polyhedron in Anton's collection is a tetrahedron.
"Cube" (without quotes), if the i-th polyhedron in Anton's collection is a cube.
"Octahedron" (without quotes), if the i-th polyhedron in Anton's collection is an octahedron.
"Dodecahedron" (without quotes), if the i-th polyhedron in Anton's collection is a dodecahedron.
"Icosahedron" (without quotes), if the i-th polyhedron in Anton's collection is an icosahedron.
#Output
Output one number — the total number of faces in all the polyhedrons in Anton's collection.


#CODE:
#include<bits/stdc++.h>
using namespace std;

int main()
{
	int n,sum=0;
	cin>>n;
	string a[n];
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	for(int i=0;i<n;i++)
	{
		if(a[i] =="Tetrahedron")
		sum+=4;
		else if(a[i]=="Cube")
		sum+=6;
		else if(a[i]=="Octahedron")
		sum+=8;
		else if(a[i]=="Dodecahedron")
		sum+=12;
		else if(a[i]=="Icosahedron")
		sum+=20;
	}
	cout<<sum;
	return 0;
}