Joseph Problem (c + + implementation)

Description: Joseph problem: There are n monkeys, in the clockwise direction of a circle King (numbering from 1 to N), from the beginning of the 1th number, counting to M, the number of the monkey out of the circle, the rest of the monkeys continue to count from 1 onwards. In this way, until there is only one monkey in the circle, this monkey is the Monkey King, programming for input n,m, output the last Monkey King number.

Input: Each line is two integers separated by a space, the first is N, the second is m (0 < M, n < 300). The last line is: 0 0

Output: For each line of input data (except the last line), the output data is also a line, that is, the last monkey's number

Input

6 2

12 4

8 3

0 0

Output

5

1

7

Analysis: Monkeys can be represented by an array, the value of the array is the number of the monkey, when a monkey out of the code to change the number to 0

Indicating that the monkey was out of the game, when the last one left is a value not 0, the value is the monkey number.

1#include <iostream>2 using namespacestd;3 4  intMain ()5 {6      intN, M;7      inta[ -];8       while(Cin >> n >> m) &&! (n = =0&& m = =0))9      {Ten           for(inti =0; I < n; i++) OneA[i] = i +1; A          intk = n;//mark the rest of the monkeys -          intj =0;//tagged m -           while(k>1) the          { -               -               for(inti =0; I < n; i++) -              { +                  if(A[i] = =0) -                      Continue; +                  Else AJ + +; at                  if(j = m)//the monkey number out of the ring becomes 0. -                  { -A[i] =0; j =0; k--; -                  } -              } -          } in           for(inti =0; I < n; i++) -          { to              if(A[i]! =0) +cout << A[i] <<Endl; -          } the      } *System"Pause"); $      return 0;Panax Notoginseng}

Joseph Problem (c + + implementation)