算法小题目小结。。。

/***** Digital Roots********/  
  
/******** written by C_Shit_Hu ************/  
  
////////////////简单大数///////////////  
  
/****************************************************************************/  
/*  
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(the length of each integer will not exceed 1000), 
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. 
*/  
/****************************************************************************/  
  
  
// 数学规律是大神！！！  
#include  
#include  
char s[1010];  
int main()  
{  
    int sum,i,len,x;  
    while(gets(s)&&s[0]!='0')  
    {  
        sum=0;  
        len=strlen(s);  
        for(i=0;i





 
  
 
 
 

/***** Hailstone HOTPO ********/  
  
/******** written by C_Shit_Hu ************/  
  
////////////////简单的数学题目///////////////  
  
/****************************************************************************/  
/*  
类似于3n+1的题目 
*/  
/****************************************************************************/  
  
  
//   
#include   
  
_int64 MAX( _int64 num )  
{  
    _int64 max =num ;  
    while(num != 1)  
    {  
        if (num%2 == 0)  
        {  
            num /=2 ;  
        }   
        else  
        {  
            num = num * 3 +1 ;  
            if (max < num)  
            {  
                max = num ;  
            }  
        }  
    }  
    return max ;  
}  
  
  
int main()  
{  
    _int64 P, order, num, i ;  
    _int64 max ;  
    scanf("%I64d", &P) ;  
    for (i=0; i