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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#429131 | #6519. X Equals Y | dnialh# | WA | 8ms | 3628kb | C++20 | 6.3kb | 2024-06-02 01:15:29 | 2024-06-02 01:15:30 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
#define F0R(i,n) for (int i = 0; i < n; i++)
#define FOR(i,a,b) for (int i = a; i <= b; i++)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define FORd(i,a,b) for (int i = (b); i >= (a); i--)
#define trav(a, x) for (auto& a : x)
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define f first
#define s second
#define mp make_pair
#define pb push_back
#define ins insert
#define lb lower_bound
#define ub upper_bound
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
const char nl = '\n';
const int MAX_N = 100011;
const ll INF = (1<<29) + 123;
const ll MOD = 1000000007; // 998244353
const ld PI = 4*atan((ld)1);
template <typename T> bool ckmin(T& a, const T& b) { return a > b ? a=b, 1 : 0; }
template <typename T> bool ckmax(T& a, const T& b) { return b > a ? a=b, 1 : 0; }
/**** Credit to chatgpt 4.0 ****/
// Stream operator for std::pair
template<typename T1, typename T2>
ostream& operator<<(ostream &out, const pair<T1, T2> &v) {
out << "(" << v.first << ", " << v.second << ")";
return out;
}
// Trait to check if a type is iterable
template<typename T, typename = void>
struct is_iterable : false_type {};
template<typename T>
struct is_iterable<T, void_t<decltype(begin(declval<T>())), decltype(end(declval<T>()))>> : true_type {};
// Stream operator for iterable types excluding std::string
template<typename TT>
typename enable_if<is_iterable<TT>::value && !is_same<TT, string>::value, ostream&>::type
operator<<(ostream& out, const TT& c) {
out << "{ ";
for (const auto& x : c) out << x << " ";
out << "}";
return out;
}
template<typename T>
ostream& operator<<(ostream& out, std::stack<T> container) {
std::vector<T> elements;
while (!container.empty()) {
elements.push_back(container.top());
container.pop();
}
std::reverse(elements.begin(), elements.end()); // Reverse to maintain order
return out << elements;
}
template<typename T>
ostream& operator<<(ostream& out, std::queue<T> container) {
std::vector<T> elements;
while (!container.empty()) {
elements.push_back(container.front());
container.pop();
}
return out << elements;
}
// Helper function to print std::priority_queue
template<typename T, typename Container, typename Compare>
ostream& operator<<(ostream& out, std::priority_queue<T, Container, Compare> pq) {
out << "{";
while (!pq.empty()) {
out << " " << pq.top();
pq.pop();
}
out << " }";
return out;
}
#ifdef DBG
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
cerr << ' ' << H;
dbg_out(T...);
}
#define dbg(...) cerr << #__VA_ARGS__ << ":", dbg_out(__VA_ARGS__);
#define dbg_array(a, n) cerr << #a << ": { "; for(int i = 0; i < n; i++) cerr << a[i] << " "; cerr << "}\n";
#else
#define dbg(...)
#define dbg_array(a, n)
#endif
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MX = 3e5+5;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t; cin >> t;
while (t--) {
int x, y, A, B; cin >> x >> y >> A >> B;
if (x > y) {
swap(x, y);
swap(A, B);
}
// brute force over all sqrt(x) and sqrt(y) possible bases
map<vi, int> v_x, v_y;
int i = 2;
for (; i*i < x; i++) {
// convert x to base i
vi cur;
int X = x;
while (X) {
cur.pb(X%i);
X /= i;
}
v_x[cur] = i;
}
int j = 2;
for (; j*j < y; j++) {
// convert y to base j
vi cur;
int Y = y;
while (Y) {
cur.pb(Y%j);
Y /= j;
}
v_y[cur] = j;
}
// check if there's a match between these two
pi ans = {-1, -1};
trav(v, v_x) {
if (v_y.count(v.f)) {
if (v.s <= A && v_y[v.f] <= B) {
ans = {v.s, v_y[v.f]};
break;
}
}
}
// now do the case where we have 1 digit between the two
if (x == y) {
int b = y+1;
assert(b >= 2);
if (b <= A && b <= B) {
ans = {b, b};
}
}
// now do the case of 2 digits
// base for x has to be in [i, A]
// base for y has to be in [j, B]
int diff = y-x;
if (diff == 0) {
ans.f = ans.s = 2;
} else {
// try all divisors
for (int k = 1; k*k <= diff; k++) {
if (diff % k) continue;
// divisors are k and diff/k
// 2nd digit is k
// a > k
// 0 < x - k*a < a
// (k+1)*a > x
// a > x/(k+1)
dbg(diff, k);
int dig = k; // dig is the "tens" digit
int a = max(x/(dig+1), dig)+1;
int b = a+diff/dig;
if (x-dig*a >= 0 && a >= i && a <= A && b >= j && b <= B) {
// ones digit is x - dig*a
int ones = x - dig*a;
ans = {a, b};
break;
}
// now try it with 2nd digit being diff/k
dig = diff/k;
a = max(x/(dig+1), dig)+1;
b = a+diff/dig;
if (x- dig*a >= 0 && a >= i && a <= A && b >= j && b <= B) {
ans = {a, b};
break;
}
}
}
// the bases are between
if (ans.f == -1) cout << "NO" << nl;
else {
cout << "YES" << nl;
assert(min(ans.f, ans.s) >= 2);
assert(ans.f <= A && ans.s <= B);
cout << ans.f << " " << ans.s << nl;
}
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 3628kb
input:
6 1 1 1000 1000 1 2 1000 1000 3 11 1000 1000 157 291 5 6 157 291 3 6 10126 114514 789 12345
output:
YES 2 2 NO YES 2 10 YES 4 5 NO YES 779 9478
result:
ok correct (6 test cases)
Test #2:
score: -100
Wrong Answer
time: 8ms
memory: 3580kb
input:
1000 920 661 756 534 52 454 31 218 980 77 812 6 729 733 289 660 161 643 21 475 602 525 329 274 782 167 279 113 875 100 388 16 426 498 341 417 433 751 312 39 91 50 47 39 941 388 247 46 725 808 148 486 945 405 700 145 647 509 152 445 45 564 16 468 843 40 530 3 722 36 323 22 568 472 443 41 38 749 25 42...
output:
YES 331 590 YES 14 148 NO YES 146 147 NO YES 66 77 YES 42 247 NO YES 214 286 NO NO NO NO YES 136 406 YES 73 96 YES 12 185 NO NO YES 28 34 YES 10 247 NO NO YES 17 193 YES 101 132 NO YES 297 701 NO YES 45 174 NO YES 28 65 NO NO YES 49 89 YES 5 6 NO NO NO NO NO YES 87 282 NO YES 157 170 YES 146 346 YES...
result:
wrong answer Integer parameter [name=b] equals to 590, violates the range [2, 534] (test case 1)