• Codeforces Round #817 (Div. 4)（6/7）

    不会有人div4的A还wa几发吧，那个人不会是我吧

    Problem - A - Codeforces

    AC代码：

    1. #include 
    2. using namespace std;
    3. using LL = long long;
    4. string a[] = {"Timur", "miurT", "Trumi", "mriTu"};
    5. void Solve() {
    6.     int n;
    7.     string s;
    8.     cin >> n >> s;
    9.     map<char, int> mp;
    10.     mp['T'] = 1;
    11.     mp['i'] = 1;
    12.     mp['m'] = 1;
    13.     mp['u'] = 1;
    14.     mp['r'] = 1;
    15.     if (n != 5) {
    16.         cout << "NO\n";
    17.         return;
    18.     }
    19.     for (int i = 0; i < n; i++) {
    20.         if (mp[s[i]] == 0) {
    21.             cout << "NO\n";
    22.             return;
    23.         } else if (mp[s[i]] == 2) {
    24.             cout << "NO\n";
    25.             return;
    26.         } else {
    27.             mp[s[i]] = 2;
    28.         }
    29.     }
    30.     cout << "YES\n";
    31. }
    32. int main(){
    33.     ios::sync_with_stdio(false);
    34.     cin.tie(nullptr);
    35.     int T;
    36.     cin >> T;
    37.     for (int i = 0; i < T; i++) {
    38.         Solve();
    39.     }
    40.     return 0;
    41. }

    Problem - B - Codeforces

    AC代码：

    1. #include 
    2. using namespace std;
    3. using LL = long long;
    4. void Solve() {
    5.     int n;
    6.     cin >> n;
    7.     string s1, s2;
    8.     cin >> s1 >> s2;
    9.     for (int i = 0; i < n; i++) {
    10.         if (s1[i] != s2[i] and (s1[i] == 'R' or s2[i] == 'R')) {
    11.             cout << "NO\n";
    12.             return;            
    13.         }
    14.     }
    15.     cout << "YES\n";
    16. }
    17. int main(){
    18.     ios::sync_with_stdio(false);
    19.     cin.tie(nullptr);
    20.     int T;
    21.     cin >> T;
    22.     for (int i = 0; i < T; i++) {
    23.         Solve();
    24.     }
    25.     return 0;
    26. }

    Problem - C - Codeforces

    AC代码：

    1. #include 
    2. using namespace std;
    3. using LL = long long;
    4. struct node {
    5.     int cnt;
    6.     bool ok1, ok2, ok3;
    7. };
    8. void Solve() {
    9.     int n;
    10.     cin >> n;
    11.     map mp;
    12.     vector<int> a(3);
    13.     for (int i = 0; i < 3; i++) {
    14.         for (int j = 0; j < n; j++) {
    15.             string s;
    16.             cin >> s;
    17.             mp[s].cnt++;
    18.             if (i == 0) {
    19.                 mp[s].ok1 = true;
    20.             } else if (i == 1) {
    21.                 mp[s].ok2 = true;
    22.             } else {
    23.                 mp[s].ok3 = true;
    24.             }
    25.         }
    26.     }
    27.     for (auto it : mp) {
    28.         if (it.second.cnt == 1) {
    29.             if (it.second.ok1) {
    30.                 a[0] += 3;
    31.             } else if (it.second.ok2) {
    32.                 a[1] += 3;
    33.             } else {
    34.                 a[2] += 3;
    35.             }
    36.         } else if (it.second.cnt == 2) {
    37.             if (it.second.ok1) {
    38.                 a[0]++;
    39.             }
    40.             if (it.second.ok2) {
    41.                 a[1]++;
    42.             }
    43.             if (it.second.ok3) {
    44.                 a[2]++;
    45.             }
    46.         }
    47.     }
    48.     for (int i = 0; i < 3; i++) {
    49.         cout << a[i] << " \n"[i == 2];
    50.     }
    51. }
    52. int main(){
    53.     ios::sync_with_stdio(false);
    54.     cin.tie(nullptr);
    55.     int T;
    56.     cin >> T;
    57.     for (int i = 0; i < T; i++) {
    58.         Solve();
    59.     }
    60.     return 0;
    61. }

    Problem - D - Codeforces

    AC代码：

    1. #include 
    2. using namespace std;
    3. using LL = long long;
    4. void Solve() {
    5.     int n;
    6.     string s;
    7.     cin >> n >> s;
    8.     int mid = n / 2;
    9.     int cnt = 0;
    10.     LL ans = 0;
    11.     vectorint, int>> a;
    12.     if (n & 1) {
    13.         for (int i = 0; i < mid; i++) {
    14.             if (s[i] == 'L') {
    15.                 cnt++;
    16.                 ans += i;
    17.                 a.push_back(make_pair(n - i - 1, i));
    18.             } else {
    19.                 ans += n - i - 1;
    20.             }
    21.         }
    22.         for (int i = mid + 1; i < n; i++) {
    23.             if (s[i] == 'R') {
    24.                 cnt++;
    25.                 ans += n - i - 1;
    26.                 a.push_back(make_pair(i, i));
    27.             } else {
    28.                 ans += i;
    29.             }
    30.         }
    31.         ans += mid;
    32.         sort(a.begin(), a.end(), [&](pair<int, int> x, pair<int, int> y) {
    33.             return x.first > y.first;
    34.         });
    35.         for (int i = 0; i < cnt; i++) {
    36.             ans += a[i].first;
    37.             if (a[i].second < mid) {
    38.                 ans -= a[i].second;
    39.             } else {
    40.                 ans -= n - a[i].second - 1;
    41.             }
    42.             cout << ans << ' ';
    43.         }
    44.         for (int i = cnt; i < n; i++) {
    45.             cout << ans << ' ';
    46.         }
    47.         cout << '\n';
    48.     } else {
    49.         for (int i = 0; i < mid; i++) {
    50.             if (s[i] == 'L') {
    51.                 cnt++;
    52.                 ans += i;
    53.                 a.push_back(make_pair(n - i - 1, i));
    54.             } else {
    55.                 ans += n - i - 1;
    56.             }
    57.         }
    58.         for (int i = mid; i < n; i++) {
    59.             if (s[i] == 'R') {
    60.                 cnt++;
    61.                 ans += n - i - 1;
    62.                 a.push_back(make_pair(i, i));
    63.             } else {
    64.                 ans += i;
    65.             }
    66.         }
    67.         sort(a.begin(), a.end(), [&](pair<int, int> x, pair<int, int> y) {
    68.             return x.first > y.first;
    69.         });
    70.         for (int i = 0; i < cnt; i++) {
    71.             ans += a[i].first;
    72.             if (a[i].second < mid) {
    73.                 ans -= a[i].second;
    74.             } else {
    75.                 ans -= n - a[i].second - 1;
    76.             }
    77.             cout << ans << ' ';
    78.         }
    79.         for (int i = cnt; i < n; i++) {
    80.             cout << ans << ' ';
    81.         }
    82.         cout << '\n';
    83.     }
    84. }
    85. int main(){
    86.     ios::sync_with_stdio(false);
    87.     cin.tie(nullptr);
    88.     int T;
    89.     cin >> T;
    90.     for (int i = 0; i < T; i++) {
    91.         Solve();
    92.     }
    93.     return 0;
    94. }

    Problem - E - Codeforces

    AC代码：

    1. #include 
    2. using namespace std;
    3. using LL = long long;
    4. LL a[1010][1010];
    5. void Solve() {
    6.     int n, q;
    7.     cin >> n >> q;
    8.     memset(a, 0, sizeof(a));
    9.     for (int i = 0; i < n; i++) {
    10.         LL x, y;
    11.         cin >> x >> y;
    12.         a[x][y] += x * y;
    13.     }
    14.     for (int i = 1; i <= 1000; i++) {
    15.         for (int j = 1; j <= 1000; j++) {
    16.             a[i][j] = a[i][j] + a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
    17.         }
    18.     }
    19.     for (int o = 0; o < q; o++) {
    20.         int hs, ws, hb, wb;
    21.         cin >> hs >> ws >> hb >> wb;
    22.         cout << a[hb - 1][wb - 1] - a[hb - 1][ws] - a[hs][wb - 1] + a[hs][ws] << '\n';
    23.     }
    24. }
    25. int main() {
    26.     ios::sync_with_stdio(false);
    27.     cin.tie(nullptr);
    28.     int T;
    29.     cin >> T;
    30.     for (int i = 0; i < T; i++) {
    31.         Solve();
    32.     }
    33.     return 0;
    34. }

    Problem - F - Codeforces

    AC代码：

    1. #include 
    2. using namespace std;
    3. using LL = long long;
    4. void Solve() {
    5.     int n, m;
    6.     cin >> n >> m;
    7.     vector a(n);
    8.     const vector<int> dirx = {1, -1, 0, 0, 1, -1, 1, -1};
    9.     const vector<int> diry = {0, 0, 1, -1, 1, 1, -1, -1};
    10.     for (int i = 0; i < n; i++) {
    11.         cin >> a[i];
    12.     }
    13.     function<bool(int, int)> check = [&](int x, int y) {
    14.         if (x < 0 or x >= n or y < 0 or y >= m) {
    15.             return false;
    16.         }
    17.         return true;
    18.     };
    19.     for (int i = 0; i < n; i++) {
    20.         for (int j = 0; j < m; j++) {
    21.             if (a[i][j] == '*') {
    22.                 int cnt = 1;
    23.                 queueint, int>> qu;
    24.                 qu.push(make_pair(i, j));
    25.                 for (int k = 0; k < 8; k++) {
    26.                     int nx = i + dirx[k], ny = j + diry[k];
    27.                     if (check(nx, ny)) {
    28.                         if (a[nx][ny] == '*') {
    29.                             cnt++;
    30.                             qu.push(make_pair(nx, ny));
    31.                         }
    32.                     }
    33.                 }
    34.                 if (cnt != 3) {
    35.                     cout << "NO\n";
    36.                     return;
    37.                 } else {
    38.                     pair<int, int> x1, x2, x3;
    39.                     x1 = qu.front();
    40.                     qu.pop();
    41.                     x2 = qu.front();
    42.                     qu.pop();
    43.                     x3 = qu.front();
    44.                     qu.pop();
    45.                     if ((x1.first == x2.first and x1.first == x3.first) or (x1.second == x2.second and x1.second == x3.second)) {
    46.                         cout << "NO\n";
    47.                         return;
    48.                     } else if ((abs(x1.first - x2.first + x1.second - x2.second) == 1 and abs(x1.first - x3.first + x1.second - x3.second) == 1) or (abs(x2.first - x1.first + x2.second - x1.second) == 1 and abs(x2.first - x3.first + x2.second - x3.second) == 1) or (abs(x3.first - x1.first + x3.second - x1.second) == 1 and abs(x3.first - x2.first + x3.second - x2.second) == 1)) {
    49.                         int sum = 1;
    50.                         for (int k = 0; k < 8; k++) {
    51.                             int nx = x1.first + dirx[k], ny = x1.second + diry[k];
    52.                             if (check(nx, ny)) {
    53.                                 if (a[nx][ny] == '*') {
    54.                                     sum++;
    55.                                 }
    56.                             }
    57.                         }
    58.                         if (sum != 3) {
    59.                             cout << "NO\n";
    60.                             return;
    61.                         }
    62.                         sum = 1;
    63.                         for (int k = 0; k < 8; k++) {
    64.                             int nx = x2.first + dirx[k], ny = x2.second + diry[k];
    65.                             if (check(nx, ny)) {
    66.                                 if (a[nx][ny] == '*') {
    67.                                     sum++;
    68.                                 }
    69.                             }
    70.                         }
    71.                         if (sum != 3) {
    72.                             cout << "NO\n";
    73.                             return;
    74.                         }
    75.                         sum = 1;
    76.                         for (int k = 0; k < 8; k++) {
    77.                             int nx = x3.first + dirx[k], ny = x3.second + diry[k];
    78.                             if (check(nx, ny)) {
    79.                                 if (a[nx][ny] == '*') {
    80.                                     sum++;
    81.                                 }
    82.                             }
    83.                         }
    84.                         if (sum != 3) {
    85.                             cout << "NO\n";
    86.                             return;
    87.                         }
    88.                         a[x1.first][x1.second] = '.';
    89.                         a[x2.first][x2.second] = '.';
    90.                         a[x3.first][x3.second] = '.';
    91.                     } else {
    92.                         cout << "NO\n";
    93.                         return;
    94.                     }
    95.                 }
    96.             }
    97.         }
    98.     }
    99.     cout << "YES\n";
    100. }
    101. int main() {
    102.     ios::sync_with_stdio(false);
    103.     cin.tie(nullptr);
    104.     int T;
    105.     cin >> T;
    106.     for (int i = 0; i < T; i++) {
    107.         Solve();
    108.     }
    109.     return 0;
    110. }

    Problem - G - Codeforces

    AC代码:

    1. #include 
    2. using namespace std;
    3. using LL = long long;
    4. void Solve() {
    5.     int n;
    6.     cin >> n;
    7.     LL c = (1ll << 31 - 1);
    8.     int l = 0, r = 0;
    9.     for (int i = 0; i < n - 2; i++) {
    10.         if (i % 2 == 0) {
    11.             l ^= i;
    12.         } else {
    13.             r ^= i;
    14.         }
    15.     }
    16.     if ((l ^ c) == (r ^ c)) {
    17.         l = 0, r = 0;
    18.         for (int i = 0; i < n - 2; i++) {
    19.             if (i % 2 == 0) {
    20.                 l ^= (i + 1);
    21.             } else {
    22.                 r ^= (i + 1);
    23.             }
    24.             cout << i + 1 << " ";
    25.         }
    26.         if ((n - 2) % 2 == 0) {
    27.             cout << (l ^ c) << " " << (r ^ c) << "\n";
    28.         } else {
    29.             cout << (r ^ c) << " " << (l ^ c) << "\n";
    30.         }
    31.     } else {
    32.         for (int i = 0; i < n - 2; i++) {
    33.             cout << i << " ";
    34.         }
    35.         if ((n - 2) % 2 == 0) {
    36.             cout << (l ^ c) << " " << (r ^ c) << "\n";
    37.         } else {
    38.             cout << (r ^ c) << " " << (l ^ c) << "\n";
    39.         }
    40.     }
    41. }
    42. int main() {
    43.     ios::sync_with_stdio(false);
    44.     cin.tie(nullptr);
    45.     int T;
    46.     cin >> T;
    47.     for (int i = 0; i < T; i++) {
    48.         Solve();
    49.     }
    50.     return 0;
    51. }

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  • 原文地址：https://blog.csdn.net/eyuhaobanga/article/details/126754401