【39.29%】【codeforces 552E】Vanya and Brackets

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

Examples

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).

【题目链接】:http://codeforces.com/contest/552/problem/E

【题解】



括号肯定要加在乘号的旁边,或者加在最开头或最结尾这样最好.

不然你加在加号的旁边没有意义的。。

因为乘号最多15个再加上开头和结尾。总共17个；

就O(17^2*长度);

枚举括号的C(N,2)个组合,然后写个函数搞出表达式的值就好.



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

string s;
vector <int> a;
stack <char> ch;
stack <LL> num;

LL cl2(LL x,LL y,char key)
{
    if (key=='+') return x+y;
    else
        return x*y;
}

void cal()
{
    LL x = num.top();
    num.pop();
    LL y = num.top();
    num.pop();
    LL z = cl2(x,y,ch.top());
    ch.pop();
    num.push(z);
}

LL cl(string s)
{
    int len = s.size();
    rep1(i,0,len-1)
        if (s[i] == '(')
            ch.push(s[i]);
        else
            if (isdigit(s[i]))
                num.push(s[i]-'0');
            else
                if (s[i]=='*')
                    ch.push(s[i]);
                else
                    if (s[i]=='+')
                    {
                        while (!ch.empty() &&ch.top()=='*')
                            cal();
                        ch.push(s[i]);
                    }
                    else
                        if (s[i]==')')
                        {
                            while (ch.top()!='(')
                                    cal();
                            ch.pop();
                        }
    while (!ch.empty())
        cal();
    return num.top();
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    cin >> s;
    a.pb(-1);
    int len = s.size();
    for (int i = 1;i <= len-1;i+=2)
        if (s[i]=='*')
            a.pb(i);
    a.pb(len);
    LL ans = 0;
    len = a.size();
    rep1(i,0,len-2)
        rep1(j,i+1,len-1)
        {
            int l = a[i],r = a[j];
            string temps = s;
            temps.insert(l+1,"(");
            temps.insert(r+1,")");
            while (!num.empty()) num.pop();
            ans = max(ans,cl(temps));
        }
    cout << ans << endl;
    return 0;
}