codewars算法习题 6级 C A Rule of Divisibility by 13

描述：When you divide the successive powers of 

10

 by 

13

 you get the following remainders of the integer divisions:

1, 10, 9, 12, 3, 4

.Then the whole pattern repeats.Hence the following method: Multiply the right most digit of the number with the left most number in the sequence shown above, the second right most digit to the second left most digit of the number in the sequence. The cycle goes on and you sum all these products. Repeat this process until the sequence of sums is stationary............................................................................Example: What is the remainder when 

1234567

 is divided by 

13

?

7×1 + 6×10 + 5×9 + 4×12 + 3×3 + 2×4 + 1×1 = 178

We repeat the process with 178:

8x1 + 7x10 + 1x9 = 87

and again with 87:

7x1 + 8x10 = 87

...........................................................................From now on the sequence is stationary and the remainder of 

1234567

 by 

13

 is the same as the remainder of 

87

 by 

13

: 

9

Call 

thirt

 the function which processes this sequence of operations on an integer 

n (>=0)

. 

thirt

 will return the stationary number.

thirt(1234567)

 calculates 178, then 87, then 87 and returns 

87

.

thirt(321)

 calculates 48, 48 and returns 

48

解析:给定一个数（long long 型），将倒着的个位数与数组1,10,9,12,3,4顺序相乘相加，数组可周而复始，相加的数重复上述过程，直至数不再变化，输出数

算法：两个函数，第一个函数先做上述过程相加的两个数的计算，然后在while循环中如果两数不相等则在传递一个数给另一个数，相等则退出，另一个函数则做上述过程中相加的计算

函数代码：long long thirt(long long n)
{
  long long transmit(long long n);
  long a,b,temp;
  a=transmit(n);
  b=transmit(a);
  if(a==b){return a;}//如果相等则直接返回
  while(a!=b){
      a=transmit(b);//大数换成第三个数
      temp=a;a=b;b=temp;//因为数会越来越小，且上一个返回数变小，所以换位使a>b，并抛弃掉第一个数
  }
  return(a);
}
long long transmit(long long n)//计算相加的函数
{
    int lsl[6]={1,10,9,12,3,4};
    int b=0,a=0,c=0;
    while(n!=0){
      a=(int)n%10;
      n=n/10;
      if(b!=6){
          c+=a*lsl[b];
          b++;
      }
      if(b==6){b=0;}
    }
    return c;

}

更简洁的方法：使用递归，一个函数足矣