CodeForces #308 Div2 E(552E Vanya and Brackets)

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Vanya is doing his maths homework. He has an expression of form $\text{[math]}$ , where x1, x2, ..., xn are digits from 1 to 9, and sign $\text{[math]}$ 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).

给定一个字符串是一个只含有加法乘法的表达式现在要在这个字符串中添加一对括号使得表达式的答案最大

枚举添加括号的位置然后计算结果就可以了

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>
#include <stack>#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)using namespace std;int Read()
{char c = getchar();while (c < '0' || c > '9') c = getchar();int x = 0;while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}return x;
}void Print(int a)
{if(a>9)Print(a/10);putchar(a%10+'0');
}vector<int> pos;
stack<ll> num;
stack<char> ch;ll cal(string s)
{while(!num.empty()) num.pop();while(!ch.empty()) ch.pop();for(int i=0;i<s.length();i++){char c=s[i];if(c>='0'&&c<='9')num.push(c-'0');else if(c=='('){ch.push(c);}else if(c==')'){while(ch.top()!='('){char x=ch.top(); ch.pop();ll a=num.top(); num.pop();ll b=num.top(); num.pop();if(x=='*')num.push(a*b);else num.push(a+b);}ch.pop();}else{if(c=='+'){while(!ch.empty()&&ch.top()=='*'){char x=ch.top(); ch.pop();ll a=num.top(); num.pop();ll b=num.top(); num.pop();if(x=='*')num.push(a*b);else num.push(a+b);}ch.push(c);}else ch.push(c);}}while(!ch.empty()){char x=ch.top(); ch.pop();ll a=num.top(); num.pop();ll b=num.top(); num.pop();if(x=='*')num.push(a*b);else num.push(a+b);}return num.top();
}int main()
{
//    fread;string str;while(cin>>str){int n=str.size();pos.clear();pos.push_back(-1);for(int i=1;i<n;i+=2)if(str[i]=='*')pos.push_back(i);pos.push_back(n);int len=pos.size();ll ans=0;for(int i=0;i<len-1;i++){for(int j=i+1;j<len;j++){string s=str;s.insert(pos[i]+1,1,'(');s.insert(pos[j]+1,1,')');//在该位置插入一个字符ans=max(ans,cal(s));}}printf("%I64d\n",ans);}return 0;
}

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