CF E. Vanya and Brackets(添加一对括号使得表达式的值最大)

E. Vanya and Brackets

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vanya is doing his maths homework. He has an expression of form

,
where x1, x2, ..., xn are
digits from 1 to 9, 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 test(s)

input

3+5*7+8*4

output

303

input

2+3*5

output

25

input

3*4*5

output

60

Note

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<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
const int N = 5005;
#define LL __int64

char fh
,s
;  //����ջ�����ʽ
LL num
;     //����ջ
int ftop,ntop ,slen;      //����ջ��������ջ��
void calculate(){
    if(fh[ftop]=='+')
        num[ntop-1]+=num[ntop],ntop--;
    else if(fh[ftop]=='-')
        num[ntop-1]-=num[ntop],ntop--;
    else if(fh[ftop]=='*')
        num[ntop-1]*=num[ntop],ntop--;
    else if(fh[ftop]=='/')
        num[ntop-1]/=num[ntop],ntop--;
    ftop--;
}
void countfuction(int l,int r){
    ftop=0;ntop=0;
    int flagNum=0;
    LL ans=0;
        for(int i=0; i<=slen; ++i){

            if(i!=slen&&(s[i]>='0'&&s[i]<='9')){
                ans=ans*10+s[i]-'0';
                flagNum=1;
            }
            else{
                if(flagNum)num[++ntop]=ans; flagNum=0;  ans=0;
                if(i==slen)break;
                if(s[i]=='+'||s[i]=='-'){
                    while(ftop&&fh[ftop]!='(') calculate();
                    fh[++ftop]=s[i];
                }
                else if(s[i]=='*'&&i==r){
                    while(ftop&&fh[ftop]!='(') calculate();   ftop--;
                    while(ftop&&(fh[ftop]=='*'||fh[ftop]=='/')) calculate();
                    fh[++ftop]=s[i];//printf("# ");
                }
                else if(s[i]=='*'||i==l){
                    while(ftop&&(fh[ftop]=='*'||fh[ftop]=='/')) calculate();
                    fh[++ftop]=s[i];
                    if(i==l)
                        fh[++ftop]='(';
                }
            }
        }
        while(ftop) calculate();

}
int main(){

    while(scanf("%s",s)>0){
        LL ans=0;
        int id[20],k=0;
        for(int i=strlen(s); i>=0; i--)
            s[i+2]=s[i];
        s[0]='1'; s[1]='*';
        slen=strlen(s);
        s[slen]='*'; s[slen+1]='1'; s[slen+2]='\0';
        slen=strlen(s);
        for(int i=0; i<slen; i++)
            if(s[i]=='*')
            id[k++]=i;

        for(int i=0; i<k-1; i++)
        for(int j=i+1; j<k; j++){
            countfuction(id[i],id[j]);
            if(num[1]>ans)
                ans=num[1];
        }
        printf("%I64d\n",ans);
    }
}