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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 Ydnialh#WA 8ms3628kbC++206.3kb2024-06-02 01:15:292024-06-02 01:15:30

Judging History

你现在查看的是最新测评结果

  • [2024-06-02 01:15:30]
  • 评测
  • 测评结果:WA
  • 用时:8ms
  • 内存:3628kb
  • [2024-06-02 01:15:29]
  • 提交

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)