Hdu1443 & poj1012: Joseph

Problem descriptionthe Joseph's problem is notoriously known. for those who are not familiar with the original problem: From among N people, numbered 1, 2 ,..., n, standing in circle every MTH is going to be executed and only the life
The last remaining person will be saved. joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. for example when n = 6 and M = 5 then the people will be executed in the order
5, 4, 6, 2, 3 and 1 will be saved.

Suppose that there are K good guys and K bad guys. in the circle the first K are good guys and the last K bad guys. you have to determine such minimal m that all the bad guys will be executed before the first good guy.

Inputthe input file consists of separate lines containing K. The last line in the input file contains 0. You can suppose that 0 <k <14.

Outputthe output file will consist of separate lines containing M corresponding to K in the input file.

Sample Input

340

 

Sample output

530

 

 

Q: k Represents k good guys and K bad guys. The first K are good guys and the last K are bad guys. According to the principle of Joseph ring game, the minimum m size when the bad guys are out and the good guys are not out

Train of Thought: due to the small value of K, the table is directly violent enumeration.

# Include <stdio. h> # include <string. h> int main () {int n, a [30], I, j, Min, Jose [14]; for (I = 1; I <14; I ++) {n = 2 * I; memset (A, 0, sizeof (a); min = 1; for (j = 1; j <= I; j ++) {A [J] = (a [J-1] + MIN-1) % (n-J + 1); // (A [J-1] + MIN-1) it calculates the positions of people that can be reached from the current position. (n-J + 1) It is the position of a person after the departure, because the number should start from him, what we get is the location of the next person out of the game. Note that this location is from 0. If (A [J] <I) // the person out of the game, J resets, min added for enumeration {J = 0; min ++ ;}} Jose [I] = min ;}while (~ Scanf ("% d", & N), n) {printf ("% d \ n", Jose [N]);} return 0 ;}