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).

给你一个表达式，只有乘号和加号，数字都是1到9，要求加一个括号，使得表达式的值最大，问最大是多少。乘号个数<=15。

左括号一定在乘号右边，右括号一定在乘号左边，因为如果不是这样的话，一定可以调整括号的位置使表达式的值增大。这个应该不难想。

于是只要枚举括号的位置然后计算表达式即可。

转载请注明出处：寻找&星空の孩子 

题目链接：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 __int64char 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;}