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  • HDU 3680 Naughty fairies BigInteger + Java +DP

    注:http://poj.org/problem?id=3278 为此题的缩水版 BFS能过
    思路:
    Step1:将数转换为二进制
    Step2:因为有-1的操作 所以不能直接贪心 通过观察可得 b是a的二进制前缀与b是二进制的前缀+1同样重要 考虑 dp[位数][状态] (状态为前缀 和前缀+1)
    Step3:建立初值 dp[i][j]=abs(a-toBase10(b的前i+1位))
    Step4:
        if(b的二进制表示第i位==0)     dp[i][0]=min(dp[i-1][0]+1,dp[i-1][1]+2)
                                    dp[i][1]=min(dp[i-1][0]+2,dp[i-1][1]+2)
        else                        dp[i][0]=min(dp[i-1][0]+2,dp[i-1][1]+2)
                                    dp[i][1]=min(dp[i-1][0]+2,dp[i-1][1]+1)
     
    代码(Java):
     
        
    import java.io.*;
    import java.math.*;
    import java.util.*;

    public class Main {
        public static BigInteger one=BigInteger.valueOf(1);
        public static BigInteger two=BigInteger.valueOf(2);
        public static String tobase2(BigInteger x){
            StringBuffer str=new StringBuffer();
            while(!x.equals(BigInteger.ZERO)){
                if(x.mod(two).equals(BigInteger.ONE)) str.append('1');
                else str.append('0');
                x=x.divide(two);
            }
            return new String(str);
        }
        public static BigInteger Min(BigInteger a,BigInteger b){
            if(a.compareTo(b)<0) return a;
            else return b;
        }
        public static BigInteger solve(BigInteger a,BigInteger b){
            String sb=tobase2(b);
            BigInteger dp[][]=new BigInteger[sb.length()][2];
            BigInteger tmp=new BigInteger("0");

            for(int i=0;i<sb.length();i++){
                tmp=tmp.multiply(two);
                if(sb.charAt(sb.length()-1-i)=='1') tmp=tmp.add(BigInteger.ONE);        
                for(int j=0;j<2;j++) {
                    dp[i][j]=a.subtract(tmp.add(BigInteger.valueOf(j))).abs();
                }
            }
            for(int i=1;i<sb.length();i++)
            {
                if(sb.charAt(sb.length()-1-i)=='0'){
                    dp[i][0]=Min(dp[i][0], Min(dp[i-1][0].add(one), dp[i-1][1].add(two)));
                    dp[i][1]=Min(dp[i][1], Min(dp[i-1][0].add(two), dp[i-1][1].add(two)));
                }else{
                    dp[i][0]=Min(dp[i][0], Min(dp[i-1][0].add(two), dp[i-1][1].add(two)));
                    dp[i][1]=Min(dp[i][1], Min(dp[i-1][0].add(two), dp[i-1][1].add(one)));
                }
            }
            return Min(dp[sb.length()-1][0], dp[sb.length()-1][1].add(one));
        }
        public static void main(String[] args) {
            Scanner cin=new Scanner(System.in);
            while(true){
                BigInteger b=cin.nextBigInteger();
                BigInteger a=cin.nextBigInteger();
                if(a.equals(b) && a.equals(BigInteger.ZERO)) break;
                if(a.compareTo(b)>=0){
                    System.out.println(a.subtract(b));
                    continue;
                }else{
                    BigInteger ans=solve(a,b);
                    System.out.println(ans);
                }
            }
            cin.close();
        }
    }
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  • 原文地址:https://www.cnblogs.com/programCaiCai/p/HDU3680.html
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