POJ2368(巴什博奕变形)

Buttons

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3073 Accepted: 781

Description

As you surely already know, Yekaterinburg has gotten its right to hold The Summer Olympic Games of the 2032. It is planned that it will be allowed to Russia as a country-organizer to emend a program of the games a bit. So, in order to improve the command result it has been decided to replace the competition in gymnastics by the competition in the new game "Buttons".
The rules of the game are very simple. There's a small heap of K buttons before two players. The players in turns take buttons from the heap, moreover, at a time one can take a number of buttons from 1 up to L. The one who takes the last button is the winner.
The rules of the Olympic Games will be a bit harder then usual. The one, who is to make a first step according to a lot, has an opportunity to fix a number K with the following restriction to it: 3 <= K <= 100 000 000 (that is the exact number of buttons that has been prepared for the Olympic tournament). The player who is to make the second step fixes a number L that satisfies the following conditions 2 <= L < K.
A very crucial task is given to your team: you are to write a program that should help the second player to make his choice. In other words, given a number K your program is to find a number L that guaranties a victory to the second player with a proper game of both sides.
So, for instance, there are only three buttons in the heap, the choice L = 2 provides for the victory of the second player. Really, if the first player takes only one button at his turn, the second one wins, taking the two last buttons. On the contrary, if the first one takes two buttons, the second one wins, taking the last button.

Input

The standard input consists of one line, which contains an only integer number K — a number of buttons in the heap, that has fixed the first player at his turn.

Output

To the standard output you are to write the only number L — the maximal number of buttons that can be taken at a time which provides for the victory of the second player. If there are several those numbers L, you should write the least. If there are no such numbers, you are to write 0 to the standard output.

Sample Input

3

Sample Output

2

题目的规则跟巴什博奕是一致的，但是问得不是谁胜谁负，给定K,求怎么尽可能小的设定L使得第二个人必胜。

注意到L>=2,我们从１开始找到其所有因子，找到最小的大于２的因子i，输出i-1即可，因为此时(m%(i-1 + 1  )==0)

注意找因子时候要用O(logN)算法

/***********************************************************> OS     : Linux 3.13.0-24-generic (Mint-17)> Author : yaolong> Mail   : [e-mail]> Time   : 2014年10月25日 星期六 11时13分18秒 **********************************************************/#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>using namespace std;int main(){    int n;    cin >> n;    int mmin = n;    for ( int  i = 1; i * i <= n; i++ )    {        if ( n % i == 0 )        {            if ( i > 2 )            {                mmin = min ( mmin, i );            }            if ( n / i > 2 )            {                mmin = min ( mmin, n / i );            }        }    }    if ( mmin > 2 )    {        cout << mmin - 1 << endl;    }    else    {        cout << 0 << endl;    }    return 0;}