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題目概要：
紀錄每個按鈕按了幾次。
解題方向：
1. 使用一個字元陣列紀錄每個英文字母指法。
2. 目前的按法跟上一個按法做比較，當是0→1時就紀錄按一次。
程式碼：
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| //Java | |
| import java.util.Scanner; | |
| class uva10415{ | |
| public static void main(String args[]){ | |
| Scanner sc=new Scanner(System.in); | |
| int cases=Integer.parseInt(sc.nextLine()); | |
| for(int i=0;i<cases;i++){ | |
| //紀錄每個英文字母所會按到的鍵。 | |
| String finger[]={"c0111001111","d0111001110","e0111001100","f0111001000","g0111000000","a0110000000","b0100000000","C0010000000","D1111001110","E1111001100","F1111001000","G1111000000","A1110000000","B1100000000"}; | |
| int count[]=new int[10]; //紀錄每個鍵所按的次數。 | |
| String st=sc.nextLine(); | |
| String current="0000000000"; //目前手指所按的鍵。 | |
| for(int j=0;j<st.length();j++){ | |
| String temp=""; | |
| //找出目前的英文字母所需要案的鍵 | |
| for(int k=0;k<finger.length;k++){ | |
| if(st.charAt(j)==finger[k].charAt(0)){ | |
| temp=finger[k].toString(); | |
| break; | |
| } | |
| } | |
| //判斷次數是否++。 | |
| StringBuilder stB=new StringBuilder(temp); | |
| temp=stB.substring(1); | |
| for(int k=0;k<10;k++){ | |
| if(current.charAt(k)!=temp.charAt(k) && temp.charAt(k)=='1') count[k]++; | |
| } | |
| current=temp; | |
| } | |
| //Output | |
| for(int j=0;j<10;j++){ | |
| if(j!=9) System.out.print(count[j]+" "); | |
| else System.out.print(count[j]); | |
| } | |
| System.out.println(""); | |
| } | |
| } | |
| } |
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