一道枚举题。
枚举左括号和右括号的位置括号，为了答案最优，左括号只能在开头或者 * 的右边。右括号只能在末尾或者 * 的左边。每一次枚举都计算一下这个加了括号后表达式的值，最后取最大值即可。
Code
#include <bits/stdc++.h>
#define ll long long
#define INF 1e9
using namespace std;
string s, backup1;
ll ans, cur;
ll facrot();
ll term();
ll expr();
ll factor() {
ll res = 0;
if (s[cur] == '(') {
cur++;
res = expr();
cur++;
}
else {
while (1) {
if (s[cur] >= '0' and s[cur] <= '9') {
res = (res * 10 + s[cur] - '0');
cur++;
}
else break;
}
}
return res;
}
ll term() {
ll res = factor();
while (1) {
if (s[cur] == '*') {
cur++;
res = res * factor();
}
else if (s[cur] == '/') {
cur++;
res = res / factor();
}
else break;
}
return res;
}
ll expr() { // 表达式求值
ll res = term();
while (1) {
if (s[cur] == '+') {
cur++;
res += term();
}
else if (s[cur] == '-') {
cur++;
res -= term();
}
else break;
}
return res;
}
signed main() {
ios :: sync_with_stdio(0);
cin >> s;
backup1 = s;
for (ll i = 0; i < s.size(); i++) {
if (s[i] == '*' or i == 0) {
if (i != 0) s.insert(i + 1, "(", 0, 1);
else s.insert(i, "(", 0, 1);
string backup2 = s;
for (ll j = i + 1; j < s.size(); j++) {
if (s[j] == '*' or j == s.size() - 1) {
cur = 0;
if (j != s.size() - 1) s.insert(j, ")", 0, 1);
else s.insert(j + 1, ")", 0, 1);
ans = max(ans, expr());
}
s = backup2;
}
}
s = backup1;
}
cout << ans;
return 0;
}