Bloomberg Codecon - The Coin Game

Introduction
You and a friend are playing a game. Being a computer scientist, and hating to lose, you write a

program that will determine ahead of time whether you will win or not.
If you will not win, you will be a sore loser and refuse to play. If you will win, you will graciously agree to play

and beat your friend.

This is likely the last game you will be playing with your friend, but in the mean time, the game rules are as follows:

There are two players: A and B.
There are N coins initially.
Players take turns picking up coins from the pile. The number of coins that any player can pick in one turn can be one of the numbers in a specific set S = {S[1], S[2], ..., S[I], ..., S[K]}. In order for a player to select S[I] coins, atleast that many coins must be remaining.
Player A starts the game.
The winner is the player who is able topick up exactly all the remaining coins.

Write a program that given N and S will determine player A's strategy that will allow him to always win, regardless of what the other player does.

If there is a strategy that will guarentee A's victory, print the minumum number of coins A must select. Input Specifications

Your program will take from STDIN

A number N ≤ 100 representing the number of coins
A number K ≤ 5 representing the size of set S
This will be followed by K lines where line I contains the value S[I] (1 ≤ I ≤ K).

Output Specifications

Based on the input,

If A has a winning strategy, print out the minimum number A can choose with their first turn and still be assured victory.
If no guaranteed winning strategy is available for A, print out -1

Sample Input/Output Input

1 1 1

Output

1

Ex planation

Player A wins by going first and taking the only coin.

Input

2 1 1

Output

-1

Ex planation

Player A can only take 1 coin and player B will then take the second coin. It is not possible for player A to win.

Input

10 3 1 3 4

Output

1

Ex planation

A picks 1 (leaving 9 coins) If B picks 1 (leaving 8 coins)

Then A picks 1 (leaving 7 coins)
If B picks 1 (leaving 6 coins) - CASE A

Then A picks 4 (leaving 2 coins)
B has to pick 1 coin, leaving 1 - A wins

If B picks 3 (leaving 4 coins) - CASE B Then A picks all 4 coins - A wins

If B picks 4 (leaving 3 coins) Then A picks all 3 - A wins

If B picks 3 (leaving 6 coins)
This reduces to CASE A and A wins

If B picks 4 (leaving 5 coins)

 

Then A picks 3 (leaving 2 coins)
Then B must pick 1, leaving 1 - A wins 

 

只需考虑先手必胜和先手必败的情况。

import java.util.*;

public class CoinGame {
	public static int minCoinNum(int n, int[] A) {
		boolean[] f = new boolean[n+1];
		int k = A.length;
		for(int i=0; i<k; i++) {
			if(A[i] > n) break;
			f[A[i]] = true;
		}
		
		for(int i=1; i<=n; i++) {
			for(int j=0; j<k; j++) {
				if(A[j] > i) break;
				f[i] = !f[i-A[j]];
				if(f[i]) break;
			}
		}
		for(int j=0; j<k; j++) {
			if(A[j] > n) break;
			if(!f[n-A[j]]) return A[j];
		}
		return -1;
	}
	
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int k = s.nextInt();
		int[] A = new int[k];
		for(int i=0; i<k; i++) {
			A[i] = s.nextInt();
		}
		s.close();
		Arrays.sort(A);
		System.out.println(minCoinNum(n, A));
	}
}