알고리즘
백준 16637번 괄호 추가하기 C++
영춘권의달인
2022. 8. 23. 10:55
괄호를 넣을 수 있는 가능한 모든 경우들을 검사하면서 가장 큰 값을 찾았다.
#include <iostream>
#include <string>
#include <vector>
#include <math.h>
#include <algorithm>
#include <utility>
#include <stack>
#include <queue>
#include <math.h>
#include <set>
#include <map>
#include <list>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <limits.h>
#include <bitset>
using namespace std;
using int64 = long long;
int n;
string str;
vector<int> compBits;
enum Operator {
Add,
Sub,
Mul,
};
map<char, Operator> oper;
int ans = INT_MIN;
void Solve(int bit, int cnt) {
if (cnt >= n / 2 + 1) {
int res = 0;
Operator currOperator = Add;
int idx = 0;
while (true) {
if (bit & compBits[idx]) {
if (idx >= n / 2) return;
int val = 0;
int left = str[idx * 2] - '0';
int right = str[idx * 2 + 2] - '0';
Operator op = oper[str[idx * 2 + 1]];
switch (op) {
case Add:
val = left + right;
break;
case Sub:
val = left - right;
break;
case Mul:
val = left * right;
break;
}
switch (currOperator) {
case Add:
res += val;
break;
case Sub:
res -= val;
break;
case Mul:
res *= val;
break;
}
if (idx + 2 > n / 2) break;
if (idx * 2 + 3 < str.length())
currOperator = oper[str[idx * 2 + 3]];
idx += 2;
}
else {
int val = str[idx * 2] - '0';
switch (currOperator) {
case Add:
res += val;
break;
case Sub:
res -= val;
break;
case Mul:
res *= val;
break;
}
if (idx >= n / 2) break;
currOperator = oper[str[idx * 2 + 1]];
idx++;
}
}
ans = max(ans, res);
return;
}
if ((bit & 1) == 1) {
Solve(bit << 1, cnt + 1);
}
else {
Solve((bit << 1) | 1, cnt + 1);
Solve(bit << 1, cnt + 1);
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> n;
cin >> str;
compBits = vector<int>(n / 2 + 1);
int compBit = 1 << n / 2;
for (int i = 0; i < compBits.size(); i++) {
compBits[i] = compBit;
compBit >>= 1;
}
oper.insert({ '+',Add });
oper.insert({ '-',Sub });
oper.insert({ '*',Mul });
Solve(0, 0);
cout << ans;
}