检查数组是否为 'right' (C++)

Question

A right array

Let's assume we have an array with an N length, made of capital letters A, B, and C. We call the array 'a right array' if between every two C letters which come one after another in the array, there are more A letters than B letters. My job is to discover whether or not a given array is 'right' and if so, I should print out "RIGHT", else I should print for how many pieces (places between to Cs) the given condition is untrue (there are more Bs than As).

Input: In the first line we enter the number of lettes in the array N (1 < N > 200). In the next line we enter the array without empty spaces in-between.

Output: Print out the answer in a single line.

Examples:

Input: 16 ABBCAABCACBAAACB Output: RIGHT
Input: 15 ABBCABCACBAAACB Output: 1
Input: 14 CABCABBCBAABBC Output: 3

现在，我已经尝试解决这个问题，但第三个示例对我不起作用 - 我得到 2 的输出，并且如上所述我应该得到 3，除此之外 - 它编译得很好。

    #include <iostream>

using namespace std;

int main()
{
    int N;
    cin >> N;
    char Ar[N];
    int A = 0;
    int B = 0;
    int piece = 0;
    int attempt = 0;

    for (int i = 0; i < N; i++) {
        cin >> Ar[i];
    }

    for (int i = 0;  i < N; i++) {
        if (Ar[i] == 'C') {
            for (int j = i + 1; i < N; j++) {
                if (Ar[j] == 'A') {
                    A++;
                } else if (Ar[j] == 'B') {
                    B++;
                } else if (Ar[j] == 'C') {
                    i = j;
                    break;
                }
            }
            if (A > B) {
                piece++;
                attempt++;
            } else if (A <= B) {
                attempt++;
            }
            A = 0;
            B = 0;
        }
    }


    if (piece == attempt) {
        cout << "RIGHT";
    } else {
        cout << attempt - piece;
    }

    return 0;
}

Answer 1

问题出在案例中

        } else if (Ar[j] == 'C') {
            i = j;
            break;
        }

原因是一旦你回到主循环i就会递增，所以结束C不会被认为是新组的开始。您的代码基本上是检查每个其他块。

你应该设置

i = j-1;

而不是，因此在递增 i 之后将是 C 的索引。

此外，您应该在评估部分时将 A 和 B 重新初始化为零。

Answer 2

您有几个问题，如下面的代码注释所述：

int N;
cin >> N;
std::vector<char> Ar(N);

for (int i = 0; i < N; i++) {
    cin >> Ar[i];
}

int piece = 0;
int attempt = 0;

for (int i = 0;  i < N - 1; i++) {
    if (Ar[i] != 'C') {
        // Skip letters until the first C
        continue;
    }
    int A = 0;
    int B = 0;
    int j = i + 1;
    for (; j < N; j++) {
        if (Ar[j] == 'A') {
            A++;
        } else if (Ar[j] == 'B') {
            B++;
        } else if (Ar[j] == 'C') {
            // We only account for blocks between Cs
            attempt++;
            if (A > B) {
                piece++;
            }
            break;
        }
    }
    // Next piece starts at j, i will be incremented by outer loop
    i = j - 1;
}

Answer 3

#include <iostream>

using namespace std;

int main() {
  int numChars;
  cin >> numChars;

  char array[numChars];

  for (int i = 0; i < numChars; ++i) {
    cin >> array[i];
  }

  int numBrokenPieces = 0;
  int numAs = 0;
  int numBs = 0;

  bool inPiece = false;
  for (int i = 0; i < numChars; ++i) {
    if (array[i] == 'C') {
      if (!inPiece) {
        inPiece = true;
        continue;
      } else {
        if (numBs >= numAs) {
          ++numBrokenPieces;
        }
        numAs = 0;
        numBs = 0;
      }
    } else {
      if (inPiece) {
        if (array[i] == 'A') {
          ++numAs;
        } else if (array[i] == 'B') {
          ++numBs;
        }
      }
    }
  }

  if (numBrokenPieces == 0) {
    cout << "RIGHT";
  } else {
    cout << numBrokenPieces;
  }

  return 0;
}

Answer 4

好吧，您也可以采用不同的方法：

string str;
bool counting = false;
int counter = 0, notRightCounter = 0;

cout << "String: ";
cin >> str;                                  // user enters whole string at once 

for (char& c : str) {                        // for each char in string
    if (c == 'C') {                          // start or stop counting when C is found
        counting = !counting;
        if (!counting && counter <= 0) {     // Check if piece between Cs is right
            notRightCounter++;
            counting = !counting;
        }
        counter = 0;
        continue;                            // Continue to next char after 'C'
    }

    if (counting)                            // Keeping count of A's and B's
        switch (c) {
            case 'A':
            counter++;
            break;
        case 'B':
            counter--;
            break;
        }
}

// Print results
if (notRightCounter != 0)                   
    cout << "Not right! " << "Not right counter: " << notRightCounter;
else
    cout << "Right!";

检查数组是否为 'right' (C++)

Checking if an array is 'right' (C++)

c++

arrays

element

char

error-checking