CodeForces - 552E Vanya and Brackets —— 加与乘运算的组合

题目链接：https://vjudge.net/contest/224393#problem/E

Vanya is doing his maths homework. He has an expression of form , where x₁, x₂, ..., x_n 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).

题意：

给出一个只有加法和乘法的算式，且数字的范围为1~9，算式里面没有括号。问：怎样加一对括号，使得算式的结果最大？

题解：
1.比划一下，可发现规律：括号加在两‘+’中间，最终结果没有改变；括号加在‘+’和‘*’之间，可使结果变大，但不一定最优。只有当括号加在两'*'之间时，结果是最大。

2.题目规定了‘*’不会超过15个，所以可直接枚举左右括号的放置位置，然后求出算式的值，取最大值即可。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <cstdlib>
 #include <string>
 #include <vector>
 #include <map>
 #include <set>
 #include <queue>
 #include <sstream>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 const double eps = 1e-;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e5+;

 char a[];
 LL s[MAXN], len;
 int pos[MAXN];

 LL solve(int l, int r)
 {
     LL sum = , t1 = , t2 = , t3 = ;
     int top = ;

     if(<=l)    //左边
     {
         s[top++] = a[]-'';
         for(int i = ; i<=l-; i+=)
         {
             if(a[i]=='*')  s[top-] = 1LL*s[top-]*(a[i+]-'');
             else if(a[i]=='+') s[top++] = (a[i+]-'');
         }
         t1 = s[--top];
         while(top) sum += s[--top];
     }

     //中间：
     s[top++] = a[l+]-'';
     for(int i = l+; i<=r-; i+=)
     {
         if(a[i]=='*')  s[top-] = 1LL*s[top-]*(a[i+]-'');
         else if(a[i]=='+') s[top++] = (a[i+]-'');
     }
     t2 = ;
     while(top) t2 += s[--top];

     // 右边：
     if(r<=len-)
     {
         s[top++] = a[r+]-'';
         for(int i = r+; i<=len-; i+=)
         {
             if(a[i]=='*')  s[top-] = 1LL*s[top-]*(a[i+]-'');
             else if(a[i]=='+') s[top++] = (a[i+]-'');
         }
         while(top>) sum += s[--top];
         t3 = s[--top];
     }
     sum += 1LL*t1*t2*t3;
     return sum;
 }

 int main()
 {
     while(scanf("%s",a+)!=EOF)
     {
         int cnt = ;
         pos[++cnt] = ;
         len = strlen(a+);
         for(int i = ; i<=len; i++)
             if(a[i]=='*') pos[++cnt] = i;
         pos[++cnt] = len+;

         LL sum = ;
         for(int l = ; l<=cnt; l++) //枚举左右括号
         for(int r = l+; r<=cnt; r++)
             sum = max(sum, solve(pos[l], pos[r]));

         cout<<sum<<endl;
     }
 }