CodeForces - 552E Vanya and Brackets —— 加与乘运算的组合

题目链接：https://vjudge.net/contest/224393#problem/E

 

Vanya is doing his maths homework. He has an expression of form , where x₁, x₂, ..., x_n 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.

Examples

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).

 

题意：

给出一个只有加法和乘法的算式，且数字的范围为1~9，算式里面没有括号。问：怎样加一对括号，使得算式的结果最大？

 

题解：
1.比划一下，可发现规律：括号加在两‘+’中间，最终结果没有改变；括号加在‘+’和‘*’之间，可使结果变大，但不一定最优。只有当括号加在两'*'之间时，结果是最大。

2.题目规定了‘*’不会超过15个，所以可直接枚举左右括号的放置位置，然后求出算式的值，取最大值即可。

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
typedef long long LL;
const double eps = 1e-8;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e5+10;

char a[10000];
LL s[MAXN], len;
int pos[MAXN];

LL solve(int l, int r)
{
    LL sum = 0, t1 = 1, t2 = 1, t3 = 1;
    int top = 0;

    if(2<=l)    //左边
    {
        s[top++] = a[1]-'0';
        for(int i = 2; i<=l-2; i+=2)
        {
            if(a[i]=='*')  s[top-1] = 1LL*s[top-1]*(a[i+1]-'0');
            else if(a[i]=='+') s[top++] = (a[i+1]-'0');
        }
        t1 = s[--top];
        while(top) sum += s[--top];
    }

    //中间：
    s[top++] = a[l+1]-'0';
    for(int i = l+2; i<=r-2; i+=2)
    {
        if(a[i]=='*')  s[top-1] = 1LL*s[top-1]*(a[i+1]-'0');
        else if(a[i]=='+') s[top++] = (a[i+1]-'0');
    }
    t2 = 0;
    while(top) t2 += s[--top];

    // 右边：
    if(r<=len-1)
    {
        s[top++] = a[r+1]-'0';
        for(int i = r+2; i<=len-1; i+=2)
        {
            if(a[i]=='*')  s[top-1] = 1LL*s[top-1]*(a[i+1]-'0');
            else if(a[i]=='+') s[top++] = (a[i+1]-'0');
        }
        while(top>1) sum += s[--top];
        t3 = s[--top];
    }
    sum += 1LL*t1*t2*t3;
    return sum;
}

int main()
{
    while(scanf("%s",a+1)!=EOF)
    {
        int cnt = 0;
        pos[++cnt] = 0;
        len = strlen(a+1);
        for(int i = 1; i<=len; i++)
            if(a[i]=='*') pos[++cnt] = i;
        pos[++cnt] = len+1;

        LL sum = 0;
        for(int l = 1; l<=cnt; l++) //枚举左右括号
        for(int r = l+1; r<=cnt; r++)
            sum = max(sum, solve(pos[l], pos[r]));

        cout<<sum<<endl;
    }
}