一个编程问题！

Question

我想设计一个程序，就是让你输入一个四位数，然后把这个四位数分成至少二分，用加减乘除连接，使得输出的值最大：比如1234，那么最大的输出值就是321*4＝1284，又如1000，最大值为100+0＝0.我想了一个晚上都没有想通，所以请大家来帮帮了。

Answer 1

import java.io.*;
import java.util.*;
public class Main
{
public static double max=Double.MIN_VALUE;
public static String rs="";
public static void main(String [] args)
{
int n=1000;
String s=String.valueOf(n);
f(s,1,0);
System.out.println(rs+"="+max);
}
private static void f(String s,int d,int fen)
{
if(d==4)
{
if(fen<1)
return;

s=s.replaceAll(" +","");
//System.out.println(s);
double dt;
try
{
dt=JiSuan.jisuan(s);
}
catch(Exception e)
{
return;
}
if(dt>=max)
{
max=dt;
rs=s;
}
return;
}

String t="";
t=s.substring(0,d*2-1)+"+"+s.substring(d*2-1);
f(t,d+1,fen+1);

t=s.substring(0,d*2-1)+"-"+s.substring(d*2-1);
f(t,d+1,fen+1);

t=s.substring(0,d*2-1)+"*"+s.substring(d*2-1);
f(t,d+1,fen+1);

t=s.substring(0,d*2-1)+"/"+s.substring(d*2-1);
f(t,d+1,fen+1);

t=s.substring(0,d*2-1)+" "+s.substring(d*2-1);
f(t,d+1,fen);
}

}class JiSuan
{
private static int index=-1;
public static double jisuan(String s) throws IOException
{
LinkedList<Token>oper=new LinkedList<Token>();
oper.addFirst(new Token('#',-1));
LinkedList<Double>num=new LinkedList<Double>();
String t="";
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)=='(')
{
if(t.equals("")!=true)
{
num.addFirst(new Double(t));
t="";
}
oper.addFirst(new Token('(',0));
}
if(s.charAt(i)==')')
{
if(t.equals("")!=true)
{
num.addFirst(new Double(t));
t="";
}
while(true)
{
Token cur=oper.removeFirst();
if(cur.c=='(')break;
double d2=num.removeFirst();
double d1=num.removeFirst();
if(cur.c=='+')
num.addFirst(d1+d2);
if(cur.c=='-')
num.addFirst(d1-d2);
if(cur.c=='*')
num.addFirst(d1*d2);
if(cur.c=='/')
{
if(d2==0)
{
throw new IOException();
}
num.addFirst(d1/d2);
}
}
}
if(s.charAt(i)=='+')
{
if(t.equals("")!=true)
{
num.addFirst(new Double(t));
t="";
}
Token tnew=new Token('+',1);
while(true)
{
Token cur=oper.removeFirst();
if(tnew.level>cur.level)
{
oper.addFirst(cur);
oper.addFirst(tnew);
break;
}
else
{
double d2=num.removeFirst();
double d1=num.removeFirst();
if(cur.c=='+')
num.addFirst(d1+d2);
if(cur.c=='-')
num.addFirst(d1-d2);
if(cur.c=='*')
num.addFirst(d1*d2);
if(cur.c=='/')
{
if(d2==0)
{
throw new IOException();
}
num.addFirst(d1/d2);
}
}
}
}
if(s.charAt(i)=='-')
{
if(t.equals("")!=true)
{
num.addFirst(new Double(t));
t="";
}
Token tnew=new Token('-',1);
while(true)
{
Token cur=oper.removeFirst();
if(tnew.level>cur.level)
{
oper.addFirst(cur);
oper.addFirst(tnew);
break;
}
else
{
double d2=num.removeFirst();
double d1=num.removeFirst();
if(cur.c=='+')
num.addFirst(d1+d2);
if(cur.c=='-')
num.addFirst(d1-d2);
if(cur.c=='*')
num.addFirst(d1*d2);
if(cur.c=='/')
{
if(d2==0)
{
throw new IOException();
}
num.addFirst(d1/d2);
}
}
}

}
if(s.charAt(i)=='*')
{
if(t.equals("")!=true)
{
num.addFirst(new Double(t));
t="";
}
Token tnew=new Token('*',2);
while(true)
{
Token cur=oper.removeFirst();
if(tnew.level>cur.level)
{
oper.addFirst(cur);
oper.addFirst(tnew);
break;
}
else
{
double d2=num.removeFirst();
double d1=num.removeFirst();
if(cur.c=='+')
num.addFirst(d1+d2);
if(cur.c=='-')
num.addFirst(d1-d2);
if(cur.c=='*')
num.addFirst(d1*d2);
if(cur.c=='/')
{
if(d2==0)
{
throw new IOException();
}
num.addFirst(d1/d2);
}
}
}

}
if(s.charAt(i)=='/')
{
if(t.equals("")!=true)
{
num.addFirst(new Double(t));
t="";
}
Token tnew=new Token('/',2);
while(true)
{
Token cur=oper.removeFirst();
if(tnew.level>cur.level)
{
oper.addFirst(cur);
oper.addFirst(tnew);
break;
}
else
{
double d2=num.removeFirst();
double d1=num.removeFirst();
if(cur.c=='+')
num.addFirst(d1+d2);
if(cur.c=='-')
num.addFirst(d1-d2);
if(cur.c=='*')
num.addFirst(d1*d2);
if(cur.c=='/')
{
if(d2==0)
{
throw new IOException();
}
num.addFirst(d1/d2);
}
}
}

}
if(s.charAt(i)>='0'&&s.charAt(i)<='9'||s.charAt(i)=='.')
t+=s.charAt(i);

}
if(t.equals("")!=true)
num.addFirst(new Double(t));
while(oper.size()>1)
{
Token cur=oper.removeFirst();
double d2=num.removeFirst();
double d1=num.removeFirst();
if(cur.c=='+')
{
num.addFirst(d1+d2);
}
if(cur.c=='-')
num.addFirst(d1-d2);
if(cur.c=='*')
num.addFirst(d1*d2);
if(cur.c=='/')
{
if(d2==0)
{
throw new IOException();
}
num.addFirst(d1/d2);
}

}

return num.getFirst();

}


}
class Token
{
public char c;
public int level;//运算优先级 ：(:0  +:1  -:1 *:2 /:2 ):3
public Token(char c,int level)
{
this.c=c;
this.level=level;
} 
public String toString()
{
return ""+c+" "+level;
}
}

Answer 2

TO: yangbc(土豆块)
哈哈，没有下多少本钱，原来写过一个计算四则表达式的类，这里直接利用了

Answer 3

TO： faen(发恩) 
先生真乃神人也.