zoukankan      html  css  js  c++  java
  • **Nim Game hard hard version

    /* In "the 100 game," two players take turns adding, to a running 
    total, any integer from 1..10. The player who first causes the running 
    total to reach or exceed 100 wins. 
    What if we change the game so that players cannot re-use integers? 
    For example, if two players might take turns drawing from a common pool of numbers 
    of 1..15 without replacement until they reach a total >= 100. This problem is 
    to write a program that determines which player would win with ideal play. 

    Write a procedure, "Boolean canIWin(int maxChoosableInteger, int desiredTotal)", 
    which returns true if the first player to move can force a win with optimal play. 

    Your priority should be programmer efficiency; don't focus on minimizing 
    either space or time complexity. 
    */ 

    Boolean canIWin(int maxChoosableInteger, int desiredTotal) { 
    // Implementation here. Write yours 

    }

    import java.util.List;
    import java.util.ArrayList;
    
    public class The100Game{
        List<Integer> pool;
        int raceTo;
        
        The100Game(int poolMax, int finalSum){
            if(finalSum > ((poolMax*poolMax + poolMax)/2)){
                throw new IllegalArgumentException("Expected sum cannot be achieved!");
            }
            
            raceTo = finalSum;
            pool = new ArrayList<Integer>();
            
            for(int i=0;i<poolMax;i++)
                pool.add(i+1);
        }
        
        boolean canIWin(){
            int turns = 0;
            while(raceTo>;0){
                turns++;
                System.out.println("Player"+( (turns%2==0)?"2":"1" )+" ==> "+pickANumber()+"   == Remaining ["+raceTo+"]");
            }
            return (turns%2==1);
        }
        
        int pickANumber(){
            int leastMax = -1;
            int len = pool.size();
            for(int i=len-1;i>=0;i--){
                int tmp = pool.get(i);
                if(tmp>=raceTo){
                    pool.remove(i);
                    raceTo -= tmp;
                    return tmp;    
                }else{
                    if(leastMax > 0){
                        if(tmp < leastMax){
                            pool.remove(i);
                            raceTo -= tmp;
                            return tmp;
                        }else{
                            continue;
                        }
                    }    
                        
                    if(i-1 >= 0) {
                        if(tmp+pool.get(i-1) < raceTo){
                            pool.remove(i);
                            raceTo -= tmp;
                            return tmp;
                        }else{
                            leastMax = raceTo - tmp;
                            i--;
                            continue;
                        }
                    }else{
                        pool.remove(i);
                        raceTo -= tmp;
                        return tmp;
                    }
                }
            }
            
            int tmp = pool.get(pool.size()-1);
            pool.remove(pool.size()-1);
            raceTo -= tmp;
            return tmp;
        }
        
        public static void main(String[] args){
            The100Game game = new The100Game(15, 100);
            System.out.println("
    Player-"+(game.canIWin()?"1":"2")+" wins!");
        }
    }

    reference:http://www.careercup.com/question?id=5116481574535168

  • 相关阅读:
    Python基础语法 第2节课(数据类型转换、运算符、字符串)
    python基础语法 第5节课 ( if 、 for )
    python基础语法 第4节课 (字典 元组 集合)
    Python基础语法 第3节课 (列表)
    A. Peter and Snow Blower 解析(思維、幾何)
    C. Dima and Salad 解析(思維、DP)
    D. Serval and Rooted Tree (樹狀DP)
    C2. Balanced Removals (Harder) (幾何、思維)
    B. Two Fairs 解析(思維、DFS、組合)
    D. Bash and a Tough Math Puzzle 解析(線段樹、數論)
  • 原文地址:https://www.cnblogs.com/hygeia/p/5162556.html
Copyright © 2011-2022 走看看