A. Anton and Polyhedrons
time limit per test:2 seconds
memory limit per test:256 megabytes
input:standard input
output:standard output
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.
Examples
Input
Copy
4
Icosahedron
Cube
Tetrahedron
Dodecahedron
Output
Copy
42
Input
Copy
3
Dodecahedron
Octahedron
Octahedron
Output
Copy
28
Note
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.
题目翻译：就是输入几个几面体，让你输出一共有几个面
题目思路：直接求就可以了，不过这里在练习用python
from sys import *
res=0
d={'T':4,'C':6,'O':8,'D':12,'I':20};
n=int(input())
for i in range(n):
s=input()
res+=d[s[0]]
print(res)
大佬代码
i=input;print(sum({'T':4,'C':6,'O':8,'D':12,'I':20}[i()[0]]for _ in range(int(i()))))
