poj2368(巴什博弈變形)
阿新 • • 發佈:2018-03-08
分享圖片 輸出 std tor pac limit total unit instance 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.
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.
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 4117 | Accepted: 1137 |
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
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
分析:題意為一堆K粒紐扣,兩個玩家輪流從堆中取紐扣,一次可以取1到L個紐扣,取得最後一個紐扣的人是贏家。
讓你找到數字最小的數字L(2<L<K)確保第二個人為贏家,如果不存在則輸出0。
很基礎的巴什博弈,只需找到K的大於3的最小因子x,那麽L=x-1。
#include<cstdio> #include<algorithm> using namespace std; int main() { int K; while(scanf("%d",&K)!=EOF) { int a[100000View Code]; int cnt=0; for(int i=1;i*i<=K;i++) { if(K%i==0) { a[cnt++]=i; a[cnt++]=K/i; } } int flag=1; sort(a,a+cnt); for(int i=0;i<cnt;i++) { if(a[i]>=3) { printf("%d\n",a[i]-1); flag=0; break; } } if(flag) printf("0\n"); } return 0; }
poj2368(巴什博弈變形)