按位贪心
核心点是如果当前为是, 并且
, 如果
但是, ,
是剩余所有位填
的结果, 必须将这些糖果消耗掉!, 也就是
个糖果需要消耗掉, 注意上取整
#include <bits/stdc++.h>
#define x first
#define y second
#define all(x) x.begin(), x.end()
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_hash(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, m;
cin >> n >> m;
LL tmp = n, ans = 0;
for (int i = 31; i >= 0; --i) {
LL v = 1ll << i;
if (tmp >= v * m) {
tmp -= v * m;
ans |= v;
}
else if (tmp > v * m - m) {
LL t = tmp - v * m + m;
tmp -= (t + v - 1) / v * v;
}
}
cout << ans << '\n';
}
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;
}

京公网安备 11010502036488号