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題目概要：
把所輸入的數字轉成2進位，並計算有多少個1。
解題方向：
Solution1: 直接使用Java提供的API(Integer.toBinaryString)，因為它回傳String，所以可以直接解析字串，並計算有幾個1。再依題目需求印出答案即可。
Solution2: 讀入數字後，自己寫一個短除法的式子，用字串儲存餘數(binary)，計算有幾個1。再依題目需求印出答案即可。
程式碼：
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| //Java-solution1 | |
| import java.util.Scanner; | |
| class uva10931{ | |
| public static void main(String args[]){ | |
| Scanner sc=new Scanner(System.in); | |
| int num; | |
| while(sc.hasNext() && (num=sc.nextInt())!=0){ | |
| String st=Integer.toBinaryString(num); | |
| //計算有多少1 | |
| int count=0; | |
| for(int i=0;i<st.length();i++){ | |
| if(st.charAt(i)=='1') count++; | |
| } | |
| //結果輸出 | |
| System.out.println("The parity of "+st+" is "+count+" (mod 2)."); | |
| } | |
| } | |
| } |
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| //Java-solution2 | |
| import java.util.Scanner; | |
| class uva10931{ | |
| public static void main(String args[]){ | |
| Scanner sc=new Scanner(System.in); | |
| int num; | |
| while((num=sc.nextInt())!=0){ | |
| //短除法，自生成binary | |
| String st=""; | |
| while(true){ | |
| if(num==1){ | |
| st="1"+st; | |
| break; | |
| } | |
| st=(char)num%2+st; | |
| num=num/2; | |
| } | |
| //計算有幾個1 | |
| int count=0; | |
| for(int i=0;i<st.length();i++){ | |
| if(st.charAt(i)=='1') count++; | |
| } | |
| //答案輸出 | |
| System.out.println("The parity of "+st+" is "+count+" (mod 2)."); | |
| } | |
| } | |
| } |
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