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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#862156#9980. Boolean Function Reconstructionucup-team2796#WA 22ms3712kbC++175.0kb2025-01-18 22:22:052025-01-18 22:22:05

Judging History

你现在查看的是最新测评结果

  • [2025-01-18 22:22:05]
  • 评测
  • 测评结果:WA
  • 用时:22ms
  • 内存:3712kb
  • [2025-01-18 22:22:05]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rrep(i, a, b) for (int i = (int)(b)-1; i >= (int)(a); i--)
#define ALL(v) (v).begin(), (v).end()
#define UNIQUE(v) sort(ALL(v)), (v).erase(unique(ALL(v)), (v).end())
#define SZ(v) (int)v.size()
#define MIN(v) *min_element(ALL(v))
#define MAX(v) *max_element(ALL(v))
#define LB(v, x) int(lower_bound(ALL(v), (x)) - (v).begin())
#define UB(v, x) int(upper_bound(ALL(v), (x)) - (v).begin())

using uint = unsigned int;
using ll = long long int;
using ull = unsigned long long;
using i128 = __int128_t;
using u128 = __uint128_t;
const int inf = 0x3fffffff;
const ll INF = 0x1fffffffffffffff;

template <typename T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T, typename U> T ceil(T x, U y) {
    assert(y != 0);
    if (y < 0)
        x = -x, y = -y;
    return (x > 0 ? (x + y - 1) / y : x / y);
}
template <typename T, typename U> T floor(T x, U y) {
    assert(y != 0);
    if (y < 0)
        x = -x, y = -y;
    return (x > 0 ? x / y : (x - y + 1) / y);
}
template <typename T> int popcnt(T x) {
    return __builtin_popcountll(x);
}
template <typename T> int topbit(T x) {
    return (x == 0 ? -1 : 63 - __builtin_clzll(x));
}
template <typename T> int lowbit(T x) {
    return (x == 0 ? -1 : __builtin_ctzll(x));
}

template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
    os << "P(" << p.first << ", " << p.second << ")";
    return os;
}
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) {
    os << "{";
    for (int i = 0; i < vec.size(); i++) {
        os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
    }
    os << "}";
    return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &map_var) {
    os << "{";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")";
        itr++;
        if (itr != map_var.end())
            os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
template <typename T> ostream &operator<<(ostream &os, const set<T> &set_var) {
    os << "{";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr;
        ++itr;
        if (itr != set_var.end())
            os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
#ifdef LOCAL
#define show(...) _show(0, #__VA_ARGS__, __VA_ARGS__)
#else
#define show(...) true
#endif
template <typename T> void _show(int i, T name) {
    cerr << '\n';
}
template <typename T1, typename T2, typename... T3>
void _show(int i, const T1 &a, const T2 &b, const T3 &...c) {
    for (; a[i] != ',' && a[i] != '\0'; i++)
        cerr << a[i];
    cerr << ":" << b << " ";
    _show(i + 1, a, c...);
}

/**
 * @brief template
 */

void Solve() {
    int N;
    cin >> N;
    vector<int> B(1<<N);
    {
        string S;
        cin >> S;
        rep(i,0,1<<N) B[i] = S[i]-'0';
    }
    auto A = B;
    {
        rep(i,0,N) {
            rep(j,0,1<<N) {
                if (j&(1<<i)) continue;
                A[j|(1<<i)] += A[j];
            }
        }
        rep(i,0,1<<N) {
            if (B[i] == 0 && A[i] > 0) {
                cout << "No" << endl;
                return;
            }
        }
    }
    if (A.back() == 0) {
        cout << "Yes" << endl;
        cout << 'F' << endl;
        return;
    }
    if (A.back() == (1<<N)) {
        cout << "Yes" << endl;
        cout << 'T' << endl;
        return;
    }
    auto DFS = [&](auto self, int Dep, int L) -> string {
        char C = 'a' + (Dep-1);
        int R = L + (1<<Dep), M = (L+R)/2;
        if (A[R-1] == 0) return "F";
        if (A[R-1] == (1<<Dep)) return "T";
        rep(i,L,M) {
            A[i|(1<<(Dep-1))] -= A[i];
        }
        string SL = self(self, Dep-1, L);
        string SR = self(self, Dep-1, M);
        if (SL == "F") {
            if (SR == "F") {
                return "F";
            }
            else if (SR == "T") {
                return string(1,C);
            }
            else {
                return '(' + SR + '&' + C + ')';
            }
        }
        else if (SL == "T") {
            return "T";
        }
        else {
            if (SR == "F") {
                return SL;
            }
            else if (SR == "T") {
                return '(' + SL + '|' + C + ')';
            }
            else {
                return '(' + SL + '|' + '(' + SR + '&' + C + ')' + ')';
            }
        }
    };
    cout << "Yes" << endl;
    cout << DFS(DFS,N,0) << endl;
}

int main() {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
int _;
cin >> _;
while(_--) {
    Solve();
}
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3712kb

input:

7
2
0001
2
0111
2
1111
3
00010111
1
10
2
0101
5
00000000000000000000000000000001

output:

Yes
(a&b)
Yes
(a|b)
Yes
T
Yes
((a&b)|((a|b)&c))
No
Yes
(a|(a&b))
Yes
((((a&b)&c)&d)&e)

result:

ok 7 lines, tightest: 4 out of 14 (7 test cases)

Test #2:

score: 0
Accepted
time: 0ms
memory: 3712kb

input:

4
1
00
1
10
1
01
1
11

output:

Yes
F
No
Yes
a
Yes
T

result:

ok 4 lines, tightest: 0 out of 11 (4 test cases)

Test #3:

score: 0
Accepted
time: 0ms
memory: 3712kb

input:

16
2
0000
2
1000
2
0100
2
1100
2
0010
2
1010
2
0110
2
1110
2
0001
2
1001
2
0101
2
1101
2
0011
2
1011
2
0111
2
1111

output:

Yes
F
No
No
No
No
No
No
No
Yes
(a&b)
No
Yes
(a|(a&b))
No
Yes
b
No
Yes
(a|b)
Yes
T

result:

ok 16 lines, tightest: 2 out of 12 (16 test cases)

Test #4:

score: 0
Accepted
time: 1ms
memory: 3584kb

input:

256
3
00000000
3
10000000
3
01000000
3
11000000
3
00100000
3
10100000
3
01100000
3
11100000
3
00010000
3
10010000
3
01010000
3
11010000
3
00110000
3
10110000
3
01110000
3
11110000
3
00001000
3
10001000
3
01001000
3
11001000
3
00101000
3
10101000
3
01101000
3
11101000
3
00011000
3
10011000
3
01011000...

output:

Yes
F
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
...

result:

ok 256 lines, tightest: 6 out of 14 (256 test cases)

Test #5:

score: 0
Accepted
time: 22ms
memory: 3584kb

input:

65536
4
0000000000000000
4
1000000000000000
4
0100000000000000
4
1100000000000000
4
0010000000000000
4
1010000000000000
4
0110000000000000
4
1110000000000000
4
0001000000000000
4
1001000000000000
4
0101000000000000
4
1101000000000000
4
0011000000000000
4
1011000000000000
4
0111000000000000
4
1111000...

output:

Yes
F
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
...

result:

ok 65536 lines, tightest: 14 out of 18 (65536 test cases)

Test #6:

score: 0
Accepted
time: 0ms
memory: 3584kb

input:

168
4
0000000000000000
4
0000000000000001
4
0000000000000011
4
0000000000000101
4
0000000000000111
4
0000000000001111
4
0000000000010001
4
0000000000010011
4
0000000000010101
4
0000000000010111
4
0000000000011111
4
0000000000110011
4
0000000000110111
4
0000000000111111
4
0000000001010101
4
000000000...

output:

Yes
F
Yes
(((a&b)&c)&d)
Yes
((b&c)&d)
Yes
(((a|(a&b))&c)&d)
Yes
(((a|b)&c)&d)
Yes
(c&d)
Yes
(((a&b)|((a&b)&c))&d)
Yes
(((a&b)|(b&c))&d)
Yes
(((a&b)|((a|(a&b))&c))&d)
Yes
(((a&b)|((a|b)&c))&d)
Yes
(((a&b)|c)&d)
Yes
((b|(b&c))&d)
Yes
((b|((a|b)&c))&d)
Yes
((b|c)&d)
Yes
(((a|(a&b))|((a|(a&b))&c))&d)
Ye...

result:

ok 168 lines, tightest: 14 out of 18 (168 test cases)

Test #7:

score: -100
Wrong Answer
time: 11ms
memory: 3584kb

input:

7581
5
00000000000000000000000000000000
5
00000000000000000000000000000001
5
00000000000000000000000000000011
5
00000000000000000000000000000101
5
00000000000000000000000000000111
5
00000000000000000000000000001111
5
00000000000000000000000000010001
5
00000000000000000000000000010011
5
0000000000000...

output:

Yes
F
Yes
((((a&b)&c)&d)&e)
Yes
(((b&c)&d)&e)
Yes
((((a|(a&b))&c)&d)&e)
Yes
((((a|b)&c)&d)&e)
Yes
((c&d)&e)
Yes
((((a&b)|((a&b)&c))&d)&e)
Yes
((((a&b)|(b&c))&d)&e)
Yes
((((a&b)|((a|(a&b))&c))&d)&e)
Yes
((((a&b)|((a|b)&c))&d)&e)
Yes
((((a&b)|c)&d)&e)
Yes
(((b|(b&c))&d)&e)
Yes
(((b|((a|b)&c))&d)&e)
Ye...

result:

wrong answer 27 operations, you can't use more than 26 operations (test case 4947)