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!
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 one number — the total number of faces in all the polyhedrons in Anton's collection.
4 Icosahedron Cube Tetrahedron Dodecahedron
42
3 Dodecahedron Octahedron Octahedron
28
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
题意:给你n个立方体,问一共有几面
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;
#define LL long long
const int INF=0x3f3f3f3f;
int main()
{
int n;
string s;
while(~scanf("%d",&n))
{
int sum=0;
for(int i=0;i<n;i++)
{
cin>>s;
if(s=="Tetrahedron") sum+=4;
else if(s=="Icosahedron") sum+=20;
else if(s=="Cube") sum+=6;
else if(s=="Octahedron") sum+=8;
else sum+=12;
}
printf("%d\n",sum);
}
return 0;
}