Anton and Polyhedrons

一、 Anton and Polyhedrons

本题链接：Anton and Polyhedrons

题目：
A. Anton and Polyhedrons
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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
4
Icosahedron
Cube
Tetrahedron
Dodecahedron
output
42

input
3
Dodecahedron
Octahedron
Octahedron
output
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.

本博客给出本题截图：

题意：不同的字符串代表不同的数字，输入n个字符串，问这些字符串对应数字的和

AC代码

#include <iostream>
#include <string>
#include <map>using namespace std;int main()
{
    map<string, int> m;m["Tetrahedron"] = 4;m["Cube"] = 6;m["Octahedron"] = 8;m["Dodecahedron"] = 12;m["Icosahedron"] = 20;int n;cin >> n;int res = 0;while (n -- ){
    string a;cin >> a;res += m[a];}cout << res << endl;return 0;
}

总结

水题，不解释