CodeForces - 552E Vanya and Brackets

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

Description

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 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).

Source

Codeforces Round #308 (Div. 2)
 
解题：暴力瞎搞，注意int溢出，傻逼逼的把栈写成int了。。。哎吸取教训
 
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 vector<int>pos;
 5 stack<LL>num;
 6 stack<char>op;
 7 LL calc(const string &str) {
 8     while(!num.empty()) num.pop();
 9     while(!op.empty()) op.pop();
10     for(int i = 0,slen = str.length(); i < slen; ++i) {
11         if(str[i] == ')') {
12             while(op.top() != '(') {
13                 LL tmp = num.top();
14                 num.pop();
15                 if(op.top() == '*') num.top() *= tmp;
16                 else if(op.top() == '+') num.top() += tmp;
17                 op.pop();
18             }
19             op.pop();
20             continue;
21         }
22         if(isdigit(str[i])) num.push(str[i] - '0');
23         else if(str[i] == '+' && !op.empty() && op.top() == '*') {
24             while(!op.empty() && op.top() == '*') {
25                 LL tmp = num.top();
26                 num.pop();
27                 num.top() *= tmp;
28                 op.pop();
29             }
30             op.push(str[i]);
31         } else op.push(str[i]);
32     }
33     while(!op.empty()) {
34         LL tmp = num.top();
35         num.pop();
36         if(op.top() == '*') num.top() *= tmp;
37         else if(op.top() == '+') num.top() += tmp;
38         op.pop();
39     }
40     return num.top();
41 }
42 int main() {
43     string str;
44     cin>>str;
45     pos.push_back(-1);
46     int slen = str.length();
47     for(int i = 1; i < slen; i += 2)
48         if(str[i] == '*') pos.push_back(i);
49     pos.push_back(slen);
50     slen = pos.size();
51     LL ret = INT_MIN;
52     for(int i = 0; i+1 < slen; ++i)
53         for(int j = i+1; j < slen; ++j) {
54             string s = str;
55             s.insert(pos[i]+1,1,'(');
56             s.insert(pos[j]+1,1,')');
57             ret = max(ret,calc(s));
58         }
59     cout<<ret<<endl;
60     return 0;
61 }
View Code

 

posted @ 2015-07-20 10:36  狂徒归来  阅读(220)  评论(0)  编辑  收藏  举报