Vanya and Brackets

Vanya and Brackets

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 552E

Description

Vanya is doing his maths homework. He has an expression of form , where x₁, x₂, ..., x_n are digits from 1 to9, and sign  represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.

Input

The first line contains expression s (1 ≤ |s| ≤ 5001, |s| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs  +  and  * .

The number of signs  *  doesn't exceed 15.

Output

In the first line print the maximum possible value of an expression.

Sample Input

Input

3+5*7+8*4

Output

303

Input

2+3*5

Output

25

Input

3*4*5

Output

60

Hint

Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.

Note to the second sample test. (2 + 3) * 5 = 25.

Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).

 

 

#include<iostream>
#include<stdio.h>
using namespace std;
char exp[50005];
int t=3;
long long  cal(int left,int right)
{
    long long  now=0;
    long long  ans=0;
    if(!left)
    {
        for(int i=1; i<t-1; i++)
        {
            if(exp[i]=='+')
            {
                if(now)
                {
                    ans+=now;
                    now=0;
                }
            }
            else if(exp[i]=='*')
            {
                now*=(exp[i+1]-'0');
                i++;
            }
            else now=exp[i]-'0';
            //cout<<now<<endl;
        }
        ans+=now;
    }
    else
    {
        long long tr=0;
        long long pos=0;
        for(int i=left+1; i<right; i++)
        {
            if(exp[i]=='+')
            {
                if(pos)
                {
                    tr+=pos;
                    pos=0;
                }
            }
            else if(exp[i]=='*')
            {
                pos*=(exp[i+1]-'0');
                i++;
            }
            else pos=exp[i]-'0';
        }
        tr+=pos;
        //cout<<tr<<endl;
        for(int i=1; i<t-1; i++)
        {
            if(exp[i]=='+')
            {
                if(now)
                {
                    ans+=now;
                    now=0;
                }
            }
            else if(exp[i]=='*')
            {
                if(i==left)
                {
                    now*=tr;
                    i=right-1;
                }
                else
                {
                    now*=(exp[i+1]-'0');
                    i++;
                }
            }
            else now=exp[i]-'0';
        }
        ans+=now;
    }
    return ans;
}
int main()
{

    while(t)
    {
        exp[t++]=getchar();
        if(exp[t-1]=='\n') break;
    }
    exp[1]='1';
    exp[2]='*';
    exp[t-1]='*';
    exp[t++]='1';
     /*for(int i=1;i<t;i++)
         cout<<exp[i]<<" ";
     //cout<<cal(0,0);*/
    long long ans=cal(0,0);
    for(int i=1; i<t-1; i++)
    {
        if(exp[i]=='*')
        {
            for(int j=i+1; j<t-1; j++)
            {
                if(exp[i]=='*')
                {
                    ans=max(ans,cal(i,j));
                }
            }
        }
    }
    printf("%I64d\n",ans);
    return 0;
}

这道题就是要想明白括号为什么要必须放在两个乘号的之间。

然后因为乘号至多有15个，所以暴力一遍就行了。