POJ-2368-Buttons [找规律]

题目链接：http://poj.org/problem?id=2368

Buttons

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

题目分析：目前有k个纽扣，两个人轮流拿石子，求出一个最小的数L，表示每次最多拿L个，（2<=L<K）使后取者能赢得比赛。若有多个则输出最小的。

如果k%(L+1)==0,则后取者赢，则只要找到K的因子中（大于2）的最小因子，让它减一即为L。 注意4！！！

#include<iostream>#include<cmath>using namespace std;int main(){long long int n,i;while(cin>>n){long long int ans=n;for(i=1;i<=sqrt(n);i++){if(n%i==0){if(i>2)ans=min(ans,i);//要求大于2if(n/i>2)//如果是4这类的情况ans=min(ans,n/i);} }if(ans>2)cout<<ans-1<<endl;elsecout<<"0"<<endl;}return 0; }