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

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).

給你一個表達式，隻有乘号和加号，數字都是1到9，要求加一個括号，使得表達式的值最大，問最大是多少。乘号個數<=15。

左括号一定在乘号右邊，右括号一定在乘号左邊，因為如果不是這樣的話，一定可以調整括号的位置使表達式的值增大。這個應該不難想。

于是隻要枚舉括号的位置然後計算表達式即可。

轉載請注明出處：尋找&星空の孩子 

題目連結：http://codeforces.com/contest/552/problem/E

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
const int N = 5005;
#define LL __int64

char fh[N],s[N];  
LL num[N];   
int ftop,ntop ,slen;     
void calculate(){
    if(fh[ftop]=='+')
        num[ntop-1]+=num[ntop],ntop--;
    else if(fh[ftop]=='-')
        num[ntop-1]-=num[ntop],ntop--;
    else if(fh[ftop]=='*')
        num[ntop-1]*=num[ntop],ntop--;
    else if(fh[ftop]=='/')
        num[ntop-1]/=num[ntop],ntop--;
    ftop--;
}
void countfuction(int l,int r){
    ftop=0;ntop=0;
    int flagNum=0;
    LL ans=0;
        for(int i=0; i<=slen; ++i){

            if(i!=slen&&(s[i]>='0'&&s[i]<='9')){
                ans=ans*10+s[i]-'0';
                flagNum=1;
            }
            else{
                if(flagNum)num[++ntop]=ans; flagNum=0;  ans=0;
                if(i==slen)break;
                if(s[i]=='+'||s[i]=='-'){
                    while(ftop&&fh[ftop]!='(') calculate();
                    fh[++ftop]=s[i];
                }
                else if(s[i]=='*'&&i==r){
                    while(ftop&&fh[ftop]!='(') calculate();   ftop--;
                    while(ftop&&(fh[ftop]=='*'||fh[ftop]=='/')) calculate();
                    fh[++ftop]=s[i];//printf("# ");
                }
                else if(s[i]=='*'||i==l){
                    while(ftop&&(fh[ftop]=='*'||fh[ftop]=='/')) calculate();
                    fh[++ftop]=s[i];
                    if(i==l)
                        fh[++ftop]='(';
                }
            }
        }
        while(ftop) calculate();

}
int main(){

    while(scanf("%s",s)>0){
        LL ans=0;
        int id[20],k=0;
        for(int i=strlen(s); i>=0; i--)
            s[i+2]=s[i];
        s[0]='1'; s[1]='*';
        slen=strlen(s);
        s[slen]='*'; s[slen+1]='1'; s[slen+2]='\0';
        slen=strlen(s);
        for(int i=0; i<slen; i++)
            if(s[i]=='*')
            id[k++]=i;

        for(int i=0; i<k-1; i++)
        for(int j=i+1; j<k; j++){
            countfuction(id[i],id[j]);
            if(num[1]>ans)
                ans=num[1];
        }
        printf("%I64d\n",ans);
    }
}
           

或着

#include <bits/stdc++.h>

using namespace std;

int n;
string str;
vector<int> pos;
stack<char> sc;
stack<long long> sn;

long long twoResult(long long a, long long b, char c) {
    return (c == '*') ? a * b : a + b;
}

void cal() {
    char t = sc.top();
    sc.pop();
    long long a = sn.top();
    sn.pop();
    long long b = sn.top();
    sn.pop();
    sn.push(twoResult(a, b, t));
}

long long expression(string s) {
    for (int i = 0; i < s.length(); i++) {
        char c = s[i];
        if (isdigit(c)) {
            sn.push(c - '0');
        } else if (c == '(') {
            sc.push(c);
        } else if (c == ')') {
            while (sc.top() != '(')
                cal();
            sc.pop();
        } else {
            if (c == '+') {
                while (!sc.empty() && sc.top() == '*')
                    cal();
                sc.push(c);
            } else sc.push(c);
        }
    }
    while (!sc.empty()) cal();
    return sn.top();
}

int main() {
    cin >> str;
    n = str.size();
    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();
    long long ans = 0;
    for (int i = 0; i < len - 1; i++) { // left
        for (int j = i + 1; j < len; j++) { // right
            string s = str;
            s.insert(pos[i] + 1, 1, '(');
            s.insert(pos[j] + 1, 1, ')');
            ans = max(ans, expression(s));
        }
    }
    cout << ans << endl;
    return 0;
}