e531: 10415 - Eb Alto Saxophone Player

UVA 10415

CPE 2020/12/22考題

解題心得：

每個音階都有對應的指法，如果欲彈奏下一個音階須按下的手指與上一個重複，則不用額外計算。

1.建立MAP來儲存每個音階要按的手指

2.建立一個狀態表(陣列)來儲存每個手指當前狀態(0沒按1按)，在建立一個記數表來記錄每個手指按下幾次。

3.每次讀到下一個音階時先去狀態表查閱當前手指狀態，如果該按而還沒按則按下(記數+1)。

4.依據MAP的資料把該放開的手指放開(狀態表)。

程式碼如下：

#include <iostream>
#include <map>
#include <vector>
#include <string>
using namespace std;

int main()
{
    cin.tie(0) , cin.sync_with_stdio(0);
    map<char , vector<int>> mc;
    mc['c'] = {0,0,1,1,1,0,0,1,1,1,1};
    mc['d'] = {0,0,1,1,1,0,0,1,1,1,0};
    mc['e'] = {0,0,1,1,1,0,0,1,1,0,0};
    mc['f'] = {0,0,1,1,1,0,0,1,0,0,0};
    mc['g'] = {0,0,1,1,1,0,0,0,0,0,0};
    mc['a'] = {0,0,1,1,0,0,0,0,0,0,0};
    mc['b'] = {0,0,1,0,0,0,0,0,0,0,0};
    mc['C'] = {0,0,0,1,0,0,0,0,0,0,0};
    mc['D'] = {0,1,1,1,1,0,0,1,1,1,0};
    mc['E'] = {0,1,1,1,1,0,0,1,1,0,0};
    mc['F'] = {0,1,1,1,1,0,0,1,0,0,0};
    mc['G'] = {0,1,1,1,1,0,0,0,0,0,0};
    mc['A'] = {0,1,1,1,0,0,0,0,0,0,0};
    mc['B'] = {0,1,1,0,0,0,0,0,0,0,0};
    int t;
    cin >> t;
    string s;
    getline(cin , s);
    while(t--){
        int finger[11] = {0};
        int fingerCount[11] = {0};
        getline(cin , s);
        for(int i = 0; i < s.length(); i++){
            if(mc.count(s[i]) == 0) continue;
            for(int j = 1; j < 11; j++) {
                if(mc[s[i]][j] == 1) {
                    if(finger[j] == 0) {
                        fingerCount[j]++;
                        finger[j] = 1;
                    }
                }
            }
            for(int k = 0; k <11; k++){
                finger[k] = mc[s[i]][k];
            }
        }
        
        for(int i = 1; i < 11 ;i++){
            cout << fingerCount[i] <<" ";
        }
        cout << "\n";
    }
    
    return 0;
}