直接枚举第一种可乐和第二种可乐的数量然后计算最大值即可
#include <bits/stdc++.h>
#define x first
#define y second
#define all(x) x.begin(), x.end()
#define vec1(T, name, n, val) vector<T> name(n, val)
#define vec2(T, name, n, m, val) vector<vector<T>> name(n, vector<T>(m, val))
#define vec3(T, name, n, m, k, val) vector<vector<vector<T>>> name(n, vector<vector<T>>(m, vector<T>(k, val)))
#define vec4(T, name, n, m, k, p, val) vector<vector<vector<vector<T>>>> name((n), vector<vector<vector<T>>>((m), vector<vector<T>>((k), vector<T>((p), (val)))))
using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PLD = pair<LD, LD>;
const int N = 1e5 + 10, MOD = 998244353;
const int INF = 1e9;
const LL LL_INF = 1e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};
istream& operator>>(istream& is, i128& val) {
string str;
is >> str;
val = 0;
bool flag = false;
if (str[0] == '-') flag = true, str = str.substr(1);
for (char& c : str) val = val * 10 + c - '0';
if (flag) val = -val;
return is;
}
ostream& operator<<(ostream& os, i128 val) {
if (val < 0) os << "-", val = -val;
if (val > 9) os << val / 10;
os << static_cast<char>(val % 10 + '0');
return os;
}
bool cmp(LD a, LD b) {
if (fabs(a - b) < EPS) return 1;
return 0;
}
struct Hash {
vector<int> h, p;
int B = 131;
Hash(const string& s) {
int n = s.size();
h.resize(n + 1, 0);
p.resize(n + 1, 1);
for (int i = 0; i < n; ++i) {
p[i + 1] = p[i] * B % MOD;
h[i + 1] = (h[i] * B + s[i]) % MOD;
}
}
LL get(int l, int r) {
LL v = h[r] - h[l - 1] * p[r - l + 1] % MOD;
v = (v % MOD + MOD) % MOD;
return v;
}
};
void solve() {
int n, a, b;
cin >> n >> a >> b;
auto calc = [&](int x, int y) {
LL cur = x + y;
while (1) {
if (x < a && y < b) break;
if (x >= a) {
int t = x / a;
cur += t;
y += t;
x %= a;
}
if (y >= b) {
int t = y / b;
cur += t;
x += t;
y %= b;
}
}
return cur;
};
LL ans = n;
for (int i = 0; i <= n; ++i) {
ans = max(ans, calc(i, n - i));
}
cout << ans << '\n';
/**/ #ifdef LOCAL
cout << flush;
/**/ #endif
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int T;
cin >> T;
while (T--) solve();
cout << fixed << setprecision(15);
return 0;
}

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