HDOJ 1013

Digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 30962    Accepted Submission(s): 9508 Problem Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit. For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39. Input The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero. Output For each integer in the input, output its digital root on a separate line of the output. Sample Input 24
39
0 Sample Output 6
3

剛開始忽略的數的大小，代碼AC

 1 #include <iostream>
 2 using namespace std;  3 void sum(int n);  4 int main()  5 {  6     int n;  7     while(cin>>n)  8  {  9         if(n == 0) 10             break; 11         else
12  { 13  sum(n); 14  } 15  } 16     return 0; 17 } 18 void sum(int n) 19 { 20     int count=0; 21     while(n>=10) 22  { 23         count+=n%10; 24         n=n/10; 25  } 26     count+=n; 27     if(count>10) 28  sum(count); 29     else
30         cout<<count<<endl; 31 }

後來改用按位來作....須要判斷是否爲0時候，要注意10這種數，最後一位爲0也須要判斷...所以改用sum是否爲0來肯定，是不是第一個數字爲0

 1 #include <iostream>
 2 using namespace std;  3 int main()  4 {  5     char c=' ';  6     int n,sum=0;  7     while(c=getchar())  8  {  9         if(sum>=10) 10                 sum=sum%10+sum/10; 11         if(c == '\n') 12  { 13             if(sum==0) 14                 break; 15             cout<<sum<<endl; 16             sum=0; 17  } 18         else
19  { 20             n=c-48; 21             sum+=n; 22  } 23  } 24     return 0; 25 }