【C# 每日一题7】求最大的结果

[pmars]

Description

Give you a expression,you can add some parenthesis to maximize result.Example as following:
1 + 2 * 3
Its result is 7 in the example.But you can add a parenthesis like (1 + 2) * 3,then the result is 9,and so it is the maximization of the example.You job is to find the maximizaton of a expression. To simplify this problem,you can assume that expression contains only operator + and *,but note that integers may be negative.

Input

Input contains multiple test cases.Each case begin with a integer N(2<=N<=100), the number of integers of a expression.Then follows the expression.Integers and operator spearate with a single whitespace.

Output

For each case print a single line with the maximized value of the expression.

Sample Input

4
1 + 2 * 3 + -1

Sample Output

8

解释一下，输入一个数n代表有下面算式里面有n个数字，每个数字中间有一个符号（+ or * ）
之后输入这个算式
在随意加括号之后求出算式可以得到的最大值！
如：
4
1 + 2 * 3 + -1
变为( 1 + 2 ) * 3 + -1 = 8
输出最大的值！！

求实现！哈哈！

[zhdi1029]

我看不明白

[hww6891]

支持一下，向牛人学习

[L19999]

学习



，，，，，，，，，

[L19999]

支持。。。。。。。。。。

[swz1222]

关注一下

[cena_jin]

太复杂了！

[acd1214]

什么啊， 给点注释

[hema_20110530]

应该是动态规划的问题

[jiangmurong]

不太懂，给点注释

[hankanling123]

支持，，，，，，，，，，

[gomoku]

[Quote=引用 47 楼 litaoye 的回复:]
粗略看了一下gomoku同志的代码，似乎没有考虑两边为负数*之后更大的情况，不知道说对了没有，没说对的话，gomoku不要介意呦。
[/Quote]
你说的是:)
下面是改正错误的代码，对几万个随机表达式（最多10多个操作数），用穷举法核实过。

void Test()
{
    int i = Maximize("-2 * 1 + 1 * -1"); // i=4
}
class Node
{
    public Node(int value, char op) {this.Value = value; this.Operator = op;}
    public int Value {get; set;}
    public char Operator {get; set;}
    public Node More { get; set; }
    public static Node operator &(Node n1, Node n2)
    {
        if (n1 == null || n2 == null) return null;
        switch(n1.Operator)
        {
            case '+': return new Node(n1.Value + n2.Value, n2.Operator );
            case '*': return new Node(n1.Value * n2.Value, n2.Operator );
            case '-': return new Node(n1.Value - n2.Value, n2.Operator );
        }
        throw new InvalidOperationException(n1.Operator + " operator not supported");
    }
}

int Maximize(string expression)
{
    string[] toks = (expression + " /").Split(' ');
    int operandCount = toks.Length / 2;

    Node[,] matrix = new Node[operandCount, operandCount];
    for (int i = 0; i < operandCount; i++)
    {
        matrix[i, i] = new Node(int.Parse(toks[i + i]), toks[i + i + 1][0]);
    }

    for (int e = 0; e < operandCount; e++)
    {
        for (int s = e - 1; s >= 0; s--)
        {
            Node max = new Node(int.MinValue, '/');
            Node min = new Node(int.MaxValue, '/');
            for (int split = e - 1; split >= s; split--)
            {
                Node[] ns = 
                {
                    matrix[s, split] & matrix[split + 1, e],
                    matrix[s, split].More & matrix[split + 1, e],
                    matrix[s, split] & matrix[split + 1, e].More,
                    matrix[s, split].More & matrix[split + 1, e].More,
                };
                foreach (Node node in ns)
                {
                    if (node != null && node.Value > max.Value) max = node;
                    if (node != null && node.Value < min.Value) min = node;
                }
            }
            matrix[s, e] = max; if(max.Value != min.Value) max.More = min;
        }
    }
    return matrix[0, matrix.GetLength(0)-1].Value;
}

[KLL]

都是牛人呀

[绿色夹克衫]

凑合写了一个

using System;
using System.Numerics;

namespace CSharpTest
{
    class Program
    {
        public static void Main()
        {
            int[] Items = new int[] { 8, -9, -1, 2, -7 };
            BigInteger[, ,] Matrix = new BigInteger[Items.Length, Items.Length, 2];

            for (int i = 0; i < Items.Length; i++)
            {
                for (int j = 0; j + i < Items.Length; j++)
                {
                    int start = j;
                    int end = i + j;

                    if (i == 0)
                        Matrix[start, end, 0] = Matrix[start, end, 1] = Items[start];
                    else
                    {
                        Matrix[start, end, 0] = int.MinValue;
                        Matrix[start, end, 1] = int.MaxValue;
                    }

                    for (int k = start; k < end; k++)
                    {
                        Matrix[start, end, 0] = Max(Matrix[start, end, 0], Matrix[start, k, 0] + Matrix[k + 1, end, 0]);
                        Matrix[start, end, 0] = Max(Matrix[start, end, 0], Matrix[start, k, 0] * Matrix[k + 1, end, 0]);
                        Matrix[start, end, 0] = Max(Matrix[start, end, 0], Matrix[start, k, 1] * Matrix[k + 1, end, 1]);

                        Matrix[start, end, 1] = Min(Matrix[start, end, 1], Matrix[start, k, 1] + Matrix[k + 1, end, 1]);
                        Matrix[start, end, 1] = Min(Matrix[start, end, 1], Matrix[start, k, 0] * Matrix[k + 1, end, 1]);
                        Matrix[start, end, 1] = Min(Matrix[start, end, 1], Matrix[start, k, 1] * Matrix[k + 1, end, 0]);
                    }
                }
            }

            Console.WriteLine(Matrix[0, Items.Length - 1, 0]);
            Console.ReadKey();
        }

        public static BigInteger Max(BigInteger a, BigInteger b)
        {
            return a > b ? a : b;
        }

        public static BigInteger Min(BigInteger a, BigInteger b)
        {
            return a < b ? a : b;
        }
    }
}

[wofeizhenlong]

路过 学习一下

[huayangniahua]

悲剧,,居然看不懂是在干嘛

[windear520]

应该以*号位中心点、、、再考虑如何让*号两边取得最大值、、、

[tiger999]

there is a easy way.

use stack

[tiger999]

1> turn the expression into reverse polish notation which can be implemented by building a expression tree and do a post order traversal.

2> use stack to evaluate the expression.

[Dj_BoAer]

好东西，学习学习

[DanaMeng]

来学习学习……