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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#723229#9463. 基础 ABC 练习题hhoppitree0 0ms0kbC++172.2kb2024-11-07 21:31:002024-11-07 21:31:01

Judging History

你现在查看的是最新测评结果

  • [2024-11-07 21:31:01]
  • 评测
  • 测评结果:0
  • 用时:0ms
  • 内存:0kb
  • [2024-11-07 21:31:00]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

const int N = 65;

unsigned int f[2][N][N][2][2];

signed main() {
    int T; scanf("%d%*d", &T);
    while (T--) {
        int n; scanf("%d", &n);
        string S1, S2; cin >> S1 >> S2;
        vector<int> o1, o2;
        for (int i = 0; i <= n; ++i) {
            if (S1[i] == '1') o1.push_back(i);
            if (S2[i] == '1') o2.push_back(i);
        }
        string S; cin >> S, S = ' ' + S;
        if (T != 0) {
            puts("-1");
            continue;
        }
        unsigned int res = 0;
        for (int i = 0; i < (int)o1.size(); ++i) {
            for (int j = 0; j < (int)o2.size(); ++j) {
                int l1 = (!i ? 0 : o1[i - 1] + 1), r1 = o1[i],
                    l2 = (!j ? 0 : o2[j - 1] + 1), r2 = o2[j], r3 = n - r1 - r2;
                memset(f[0], 0, sizeof(f[0]));
                f[0][0][0][0 >= l1][0 >= l2] = 1;
                for (int k = 1; k <= n * 3; ++k) {
                    memset(f[k & 1], 0, sizeof(f[k & 1]));
                    for (int p = 0; p <= n; ++p) {
                        for (int q = 0; q <= n; ++q) {
                            int r = k - 1 - p - q;
                            if (r < 0 || r > n) continue;
                            for (int a = 0; a < 2; ++a) {
                                for (int b = 0; b < 2; ++b) {
                                    for (int c = 0; c < 3; ++c) {
                                        if ((S[k] != '?' && S[k] != "ABC"[c]) || !f[(k & 1) ^ 1][p][q][a][b]) continue;
                                        int tp = p + (c == 0), tq = q + (c == 1), tr = r + (c == 2);
                                        if (tp - tr <= r1 && tq - tp <= r2 && tr - tq <= r3) {
                                            f[k & 1][tp][tq][a || (tp - tr >= l1)][b || (tq - tp >= l2)] += f[(k & 1) ^ 1][p][q][a][b];
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
                res += f[n & 1][n][n][1][1];
            }
        }
        printf("%u\n", res);
    }
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Subtask #1:

score: 0
Time Limit Exceeded

Test #1:

score: 0
Time Limit Exceeded

input:

60 1
1
11
11
ABC
2
111
111
CABABC
3
1111
1111
CAABBCBAC
4
11111
11111
BACBBACBACAC
5
111111
111111
CABCCBBAABCCBAA
6
1111111
1111111
ABABABCACBCBCCACBA
7
11111111
11111111
BCAABACBBCBBABCCAACAC
8
111111111
111111111
CCBCBBBCAABCBCAAAAACBCBA
9
1111111111
1111111111
CCCCACABCBABAABCCAABABBCBBA
10
1111...

output:

-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1

result:


Subtask #2:

score: 0
Skipped

Dependency #1:

0%

Subtask #3:

score: 0
Time Limit Exceeded

Test #22:

score: 0
Time Limit Exceeded

input:

60 3
1
11
11
???
2
111
111
??????
3
1111
1111
?????????
4
11111
11111
????????????
5
111111
111111
???????????????
6
1111111
1111111
??????????????????
7
11111111
11111111
?????????????????????
8
111111111
111111111
????????????????????????
9
1111111111
1111111111
???????????????????????????
10
1111...

output:

-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1

result:


Subtask #4:

score: 0
Skipped

Dependency #1:

0%

Subtask #5:

score: 0
Skipped

Dependency #1:

0%