Codeforces Round #308 (Div. 2) E. Vanya and Brackets

E. Vanya and Brackets

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard 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.

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).

题目要求对一个只有加号乘号的数式加一对括号，要求和最大是多少，*的个数不超过15个，这个问题如果枚举至少也o(n^3),所以复杂度太高，考虑到

*号的个数很少，就可以想到（一定在*号的右边或开始，）一定在*的左边或开始，因为，如果存在这样的式子+(...)+,这括号加的没有意义，不能使等式

变大，如果存在*（....)+这样的式子，我们把）向右移可使式子变大，如果存在+(...)*,我们可以把（向左移使得式子变大，这样，枚举一下，总的复杂度为n*17*17个，足够了！

#define INF 9000000000000000000
#define EPS (double)1e-9
#define mod 1000000007
#define PI 3.14159265358979
//*******************************************************************************/
#endif
#define N 5005
#define MOD 1000000007
int n,len,leftt[30],leftnum,rightt[30],rightnum;
long long ans;
char str
,temp
;
stack<char> opS;
stack<long long> numS;
int getNum(char c){
if(c == '(')return 3;
else if(c == '*')return 2;
else if(c == '+')return 1;
else if(c == ')')return 0;
return -1;
}
bool isdigit(char c){
if(c>='0' && c<='9')return true;
return false;
}
bool PopCal(){
char op = opS.top();opS.pop();
//cout<<op<<endl;
if(op == '+'){
long long num1 = numS.top();numS.pop();
long long num2 = numS.top();numS.pop();
numS.push(num1 + num2);
}
else if(op == '*'){
long long num1 = numS.top();numS.pop();
long long num2 = numS.top();numS.pop();
numS.push(num1 * num2);
}
else
return true;
return false;
}
long long cal(char s[],int l){
while(!opS.empty()) opS.pop();
while(!numS.empty()) numS.pop();
for(int i=0;i<l;i++){
if(isdigit(s[i]))
numS.push(s[i]-'0');
else {
while(!opS.empty() && getNum(s[i]) <= getNum(opS.top())){
if(opS.top() == '(' && s[i]!=')')break;
if(PopCal())
break;
}
if(s[i] != ')')
opS.push(s[i]);
}
}
while(!opS.empty()){
PopCal();
}
return numS.top();
}
int main()
{
//cout<<cal("3+5*(7+8)*4",11);
while (SS(str) != EOF)
{
len = strlen(str);leftnum = 0;rightnum = 0;leftt[leftnum++] = -1;ans = 0;
FI(len){
if(str[i] == '*'){
leftt[leftnum++] = i;
rightt[rightnum++] = i;
}
}
rightt[rightnum++] = len;
FI(leftnum)
FJ(rightnum){
if(rightt[j] > leftt[i]){
int count = 0;
if(leftt[i] == -1) temp[count++] = '(';
for(int k=0;k<len;k++){
if(k == leftt[i]){
temp[count++] = str[k];
temp[count++] = '(';
}
else if(k == rightt[j]){
temp[count++] = ')';
temp[count++] = str[k];
}
else {
temp[count++] = str[k];
}
}
if(rightt[j] == len) temp[count++] = ')';
temp[count] = '\0';
//printf("%d %d %d %d %s ",i,j,leftt[i],rightt[j],temp);
ans = max(ans,cal(temp,len+2));
//cout<<cal(temp,len+2)<<endl;
}
}
cout<<ans<<endl;
}
return 0;
}