| 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 }