题目链接
讲道理距离上一次写这种求值的题已经不知道多久了。
括号肯定是左括号在乘号的右边， 右括号在左边。 否则没有意义。 题目说乘号只有15个， 所以我们枚举就好了。
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
stack <char> sign;
stack <ll> digit;
int a[20];
string str;
void cal() {
char s = sign.top(); sign.pop();
ll x = digit.top(); digit.pop();
ll y = digit.top(); digit.pop();
ll tmp;
if(s == '+')
tmp = x+y;
else
tmp = x*y;
digit.push(tmp);
}
ll solve(){
if(!digit.empty())
digit.pop();
for(int i = 0; i < str.size(); i ++) {
if(isdigit(str[i])) {
digit.push(str[i]-'0');
} else if(str[i] == '(') {
sign.push('(');
} else if(str[i] == ')') {
while(sign.top() != '(') {
cal();
}
sign.pop();
} else if(str[i] == '*') {
sign.push('*');
} else {
while(!sign.empty() && sign.top() == '*')
cal();
sign.push('+');
}
}
while(!sign.empty())
cal();
return digit.top();
}
int main()
{
string s;
cin>>s;
int cnt = 0;
a[cnt++] = -1;
for(int i = 0; i < s.size(); i++)
if(s[i] == '*')
a[cnt++] = i;
a[cnt++] = s.size();
ll ans = 0;
for(int i = 0; i < cnt; i ++) {
for(int j = i+1; j < cnt; j ++) {
str = s;
str.insert(a[i]+1, 1, '(');
str.insert(a[j]+1, 1, ')');
ans = max(ans, solve());
}
}
cout<<ans<<endl;
return 0;
}