母音と子音の位置関係を変えない単語の配列？-C++-php.cn

母音と子音の位置関係を変えない単語の配列？

n 個の要素 (n

方法は非常に簡単です。与えられた文字列内の母音と子音の数を数え、次に母音のみを配置できる方法の数を見つけ、次に子音のみを配置できる方法の数を見つけて、これら 2 つの結果を乗算する必要があります。ウェイの合計数を取得します。

アルゴリズム

arrangeWayCount(str)

Begin define an array ‘freq’ to store frequency. count and place frequency of each characters in freq array. such that freq[‘0’] will hold frequency of letter ‘a’, freq[1] will hold frequency of ‘b’ and so on. v := number of vowels, and c := number of consonants in str vArrange := factorial of v for each vowel v in [a, e, i, o, u], do vArrange := vArrange / factorial of the frequency of v done cArrange := factorial of c for each consonant con, do cArrange := cArrange / factorial of the frequency of con done return vArrange * cArrange End

ログイン後にコピー

例

#include  using namespace std; long long factorial(int n){ if(n == 0 || n == 1) return 1; return n*factorial(n-1); } long long arrangeWayCount(string str){ long long freq[27] = {0}; //fill frequency array to 0 int v = 0, c = 0; for (int i = 0; i < str.length(); i++) { freq[str[i] - 'a']++; if (str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u') { v++; }else c++; } long long arrangeVowel; arrangeVowel = factorial(v); arrangeVowel /= factorial(freq[0]); // vowel a arrangeVowel /= factorial(freq[4]); // vowel e arrangeVowel /= factorial(freq[8]); // vowel i arrangeVowel /= factorial(freq[14]); // vowel o arrangeVowel /= factorial(freq[20]); // vowel u long long arrangeConsonant; arrangeConsonant = factorial(c); for (int i = 0; i < 26; i++) { if (i != 0 && i != 4 && i != 8 && i != 14 && i != 20) arrangeConsonant /= factorial(freq[i]); //frequency of all characters except vowels } long long total = arrangeVowel * arrangeConsonant; return total; } main() { string str = "computer"; long long ans = arrangeWayCount(str); cout << "Possible ways to arrange: " << ans << endl; }

ログイン後にコピー