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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
#120125#6128. Flippy Sequencehos_lyricAC ✓109ms5584kbC++141.9kb2023-07-06 14:06:032023-07-06 14:06:04

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-07-06 14:06:04]
  • 评测
  • 测评结果:AC
  • 用时:109ms
  • 内存:5584kb
  • [2023-07-06 14:06:03]
  • 提交

answer

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }


int N;
char S[1'000'010];
char T[1'000'010];

int main() {
  for (int numCases; ~scanf("%d", &numCases); ) { for (int caseId = 1; caseId <= numCases; ++caseId) {
    scanf("%d", &N);
    scanf("%s", S);
    scanf("%s", T);
    
    vector<pair<int, int>> ps;
    for (int i = 0, j; i < N; i = j) {
      for (j = i; j < N && (S[i] != T[i]) == (S[j] != T[j]); ++j) {}
      if (S[i] != T[i]) {
        ps.emplace_back(i, j);
      }
    }
    const int psLen = ps.size();
// cerr<<"ps = "<<ps<<endl;
    
    Int ans = 0;
    if (psLen == 0) {
      ans = 1LL * N * (N + 1) / 2;
    } else if (psLen == 1) {
      ans = 2 * (N - 1);
    } else if (psLen == 2) {
      ans = 6;
    }
    printf("%lld\n", ans);
  }
#ifndef LOCAL
  break;
#endif
  }
  return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3688kb

input:

3
1
1
0
2
00
11
5
01010
00111

output:

0
2
6

result:

ok 3 number(s): "0 2 6"

Test #2:

score: 0
Accepted
time: 109ms
memory: 5584kb

input:

126648
1
0
0
1
1
0
2
01
01
2
01
11
2
10
11
2
11
00
3
011
011
3
010
110
3
011
001
3
111
001
3
001
000
3
101
000
3
011
000
3
111
000
4
1111
1111
4
1110
0110
4
0010
0110
4
1011
0111
4
1001
1011
4
0100
1110
4
0000
0110
4
0111
1001
4
1001
1000
4
1011
0010
4
0001
0100
4
1000
0101
4
0100
0111
4
1101
0110
4...

output:

1
0
3
2
2
2
6
4
4
4
4
6
4
4
10
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
15
8
8
8
8
6
8
8
8
6
6
6
8
6
8
8
8
6
6
6
6
0
6
6
8
6
6
6
8
6
8
8
21
10
10
10
10
6
10
10
10
6
6
6
10
6
10
10
10
6
6
6
6
0
6
6
10
6
6
6
10
6
10
10
10
6
6
6
6
0
6
6
6
0
0
0
6
0
6
6
10
6
6
6
6
0
6
6
10
6
6
6
10
6
10
10
28
12
12
12
12
6
12
12
1...

result:

ok 126648 numbers