给定一个字符串W，更改该字符串，使其不包含任何作为其子字符串之一的“禁止”字符串S1到Sn。管理这一变化的规则如下： 1.字符大小写无关紧要，即“XyZ”与“XyZ”相同。 2.更改原始字符串中子字符串包含的所有字符 W 这样一封信 书信电报 在更改的字符串中发生的最大次数，并且更改的字符串是从所有可能的组合中可以获得的最小字符串。 3.不在禁止子字符串内的字符不允许更改。 4.更改字符的大小写必须与字符串中该位置的原始字符的大小写相同 W . 5.如果这封信 书信电报 是子字符串的一部分，则必须根据上述规则将其更改为其他字符。 例如：
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Input : n = 3 s1 = "etr" s2 = "ed" s3 = "ied" W = "PEtrUnited" letter = "d"Output : PDddUnitdaInput : n = 1 s1 = "PetrsDreamOh" W = "PetrsDreamOh" letter = hOutput : HhhhhHhhhhHa
说明： 例1： 第一个字符P不属于任何子字符串，因此不会更改。接下来的三个字符构成子字符串“etr”，并更改为“Ddd”。接下来的四个字符不属于任何禁止的子字符串，并且保持不变。接下来的两个字符构成子字符串“ed”，并被更改为“da”，因为d本身是最后一个字符，所以它被另一个字符“a”替换，这样字符串在词典中是最小的。 注意： “Etr”=“Etr”和更改后的子字符串“Ddd”的第一个字符为“D”，因为“Etr”的第一个字母是大写的。 例2： 由于整个字符串都是禁止的字符串，因此根据大小写将其替换为字母“h”，从第一个字符到最后第二个字符。最后一个字符是“h”，因此它被字母“a”替换，这样字符串在词典中是最小的。
方法： 检查字符串W的每个字符，看它是否是任何子字符串的开头。使用另一个数组来记录W的字符，这些字符是禁止字符串的一部分，需要更改。 根据以下规则更改字符： 1.如果W[i]=字母，字母=’a’，那么W[i]=’b’ 2.如果W[i]=字母和字母！=’然后W[i]=“a” 3.如果W[i]！=字母W[i]=字母 当W[i]是大写字符时，也将使用第一和第二个条件。 以下是上述方法的实施情况：
C++
// CPP program to remove the forbidden strings #include <bits/stdc++.h> using namespace std; // pre[] keeps record of the characters // of w that need to be changed bool pre[100]; // number of forbidden strings int n; // given string string w; // stores the forbidden strings string s[110]; // letter to replace and occur // max number of times char letter; // Function to check if the particula // r substring is present in w at position void verify( int position, int index) { int l = w.length(); int k = s[index].length(); // If length of substring from this // position is greater than length // of w then return if (position + k > l) return ; bool same = true ; for ( int i = position; i < position + k; i++) { // n and n1 are used to check for // substring without considering // the case of the letters in w // by comparing the difference // of ASCII values int n, n1; char ch = w[i]; char ch1 = s[index][i - position]; if (ch >= 'a' && ch <= 'z' ) n = ch - 'a' ; else n = ch - 'A' ; if (ch1 >= 'a' && ch1 <= 'z' ) n1 = ch1 - 'a' ; else n1 = ch1 - 'A' ; if (n != n1) same = false ; } // If same == true then it means a // substring was found starting at // position therefore all characters // from position to length of substring // found need to be changed therefore // they needs to be marked if (same == true ) { for ( int i = position; i < position + k; i++) pre[i] = true ; return ; } } // Function implementing logic void solve() { int l = w.length(); int p = letter - 'a' ; for ( int i = 0; i < 100; i++) pre[i] = false ; // To verify if any substring is // starting from index i for ( int i = 0; i < l; i++) { for ( int j = 0; j < n; j++) verify(i, j); } // Modifying the string w // according to th rules for ( int i = 0; i < l; i++) { if (pre[i] == true ) { if (w[i] == letter) w[i] = (letter == 'a' ) ? 'b' : 'a' ; // This condition checks // if w[i]=upper(letter) else if (w[i] == 'A' + p) w[i] = (letter == 'a' ) ? 'B' : 'A' ; // This condition checks if w[i] // is any lowercase letter apart // from letter. If true replace // it with letter else if (w[i] >= 'a' && w[i] <= 'z' ) w[i] = letter; // This condition checks if w[i] // is any uppercase letter apart // from letter. If true then // replace it with upper(letter). else if (w[i] >= 'A' && w[i] <= 'Z' ) w[i] = 'A' + p; } } cout << w; } // Driver function for the program int main() { n = 3; s[0] = "etr" ; s[1] = "ed" ; s[2] = "ied" ; w = "PEtrUnited" ; letter = 'd' ; // Calling function solve(); return 0; } |
JAVA
// Java program to remove forbidden strings import java.io.*; class rtf { // number of forbidden strings public static int n; // original string public static String z; // forbidden strings public static String s[] = new String[ 100 ]; // to store original string // as character array public static char w[]; // letter to replace and occur // max number of times public static char letter; // pre[] keeps record of the characters // of w that need to be changed public static boolean pre[] = new boolean [ 100 ]; // Function to check if the particular // substring is present in w at position public static void verify( int position, int index) { int l = z.length(); int k = s[index].length(); // If length of substring from this // position is greater than length // of w then return if (position + k > l) return ; boolean same = true ; for ( int i = position; i < position + k; i++) { // n and n1 are used to check for // substring without considering // the case of the letters in w // by comparing the difference // of ASCII values int n, n1; char ch = w[i]; char ch1 = s[index].charAt(i - position); if (ch >= 'a' && ch <= 'z' ) n = ch - 'a' ; else n = ch - 'A' ; if (ch1 >= 'a' && ch1 <= 'z' ) n1 = ch1 - 'a' ; else n1 = ch1 - 'A' ; if (n != n1) same = false ; } // If same == true then it means a substring // was found starting at position therefore // all characters from position to length // of substring found need to be changed // therefore they need to be marked if (same == true ) { for ( int i = position; i < position + k; i++) pre[i] = true ; return ; } } // Function performing calculations. public static void solve() { w = z.toCharArray(); letter = 'd' ; int l = z.length(); int p = letter - 'a' ; for ( int i = 0 ; i < 100 ; i++) pre[i] = false ; // To verify if any substring is // starting from index i for ( int i = 0 ; i < l; i++) { for ( int j = 0 ; j < n; j++) verify(i, j); } // Modifying the string w // according to th rules for ( int i = 0 ; i < l; i++) { if (pre[i] == true ) { if (w[i] == letter) w[i] = (letter == 'a' ) ? 'b' : 'a' ; // This condition checks // if w[i]=upper(letter) else if (w[i] == ( char )(( int ) 'A' + p)) w[i] = (letter == 'a' ) ? 'B' : 'A' ; // This condition checks if w[i] // is any lowercase letter apart // from letter. If true replace // it with letter else if (w[i] >= 'a' && w[i] <= 'z' ) w[i] = letter; // This condition checks if w[i] // is any uppercase letter apart // from letter. If true then // replace it with upper(letter). else if (w[i] >= 'A' && w[i] <= 'Z' ) w[i] = ( char )(( int ) 'A' + p); } } System.out.println(w); } // Driver function for the program public static void main(String args[]) { n = 3 ; s[ 0 ] = "etr" ; s[ 1 ] = "ed" ; s[ 2 ] = "ied" ; z = "PEtrUnited" ; solve(); } } |
Python3
# Python program to remove the forbidden strings # pre[] keeps record of the characters # of w that need to be changed pre = [ False ] * 100 # stores the forbidden strings s = [ None ] * 110 # Function to check if the particula # r substring is present in w at position def verify(position, index): l = len (w) k = len (s[index]) # If length of substring from this # position is greater than length # of w then return if (position + k > l): return same = True for i in range (position, position + k): # n and n1 are used to check for # substring without considering # the case of the letters in w # by comparing the difference # of ASCII values ch = w[i] ch1 = s[index][i - position] if (ch > = 'a' and ch < = 'z' ): n = ord (ch) - ord ( 'a' ) else : n = ord (ch) - ord ( 'A' ) if (ch1 > = 'a' and ch1 < = 'z' ): n1 = ord (ch1) - ord ( 'a' ) else : n1 = ord (ch1) - ord ( 'A' ) if (n ! = n1): same = False # If same == true then it means a # substring was found starting at # position therefore all characters # from position to length of substring # found need to be changed therefore # they needs to be marked if (same = = True ): for i in range ( position, position + k): pre[i] = True return # Function implementing logic def solve(): l = len (w) p = ord (letter) - ord ( 'a' ) # To verify if any substring is # starting from index i for i in range (l): for j in range (n): verify(i, j) # Modifying the string w # according to th rules for i in range (l): if (pre[i] = = True ): if (w[i] = = letter): w[i] = 'b' if (letter = = 'a' ) else 'a' # This condition checks # if w[i]=upper(letter) elif (w[i] = = str ( ord ( 'A' ) + p)): w[i] = 'B' if (letter = = 'a' ) else 'A' # This condition checks if w[i] # is any lowercase letter apart # from letter. If true replace # it with letter elif (w[i] > = 'a' and w[i] < = 'z' ): w[i] = letter # This condition checks if w[i] # is any uppercase letter apart # from letter. If true then # replace it with upper(letter). elif (w[i] > = 'A' and w[i] < = 'Z' ): w[i] = chr ( ord ( 'A' ) + p) print (''.join(w)) # Driver function for the program # number of forbidden strings n = 3 s[ 0 ] = "etr" s[ 1 ] = "ed" s[ 2 ] = "ied" # given string w = "PEtrUnited" w = list (w) # letter to replace and occur # max number of times letter = 'd' # Calling function solve() # This code is contributed by ankush_953 |
C#
// C# program to remove forbidden strings using System; class GFG { // number of forbidden strings public static int n; // original string public static string z; // forbidden strings public static string [] s = new string [100]; // to store original string // as character array public static char [] w; // letter to replace and occur // max number of times public static char letter; // pre[] keeps record of the characters // of w that need to be changed public static bool [] pre = new bool [100]; // Function to check if the particular // substring is present in w at position public static void verify( int position, int index) { int l = z.Length; int k = s[index].Length; // If length of substring from this // position is greater than length // of w then return if (position + k > l) { return ; } bool same = true ; for ( int i = position; i < position + k; i++) { // n and n1 are used to check for // substring without considering // the case of the letters in w // by comparing the difference // of ASCII values int n, n1; char ch = w[i]; char ch1 = s[index][i - position]; if (ch >= 'a' && ch <= 'z' ) { n = ch - 'a' ; } else { n = ch - 'A' ; } if (ch1 >= 'a' && ch1 <= 'z' ) { n1 = ch1 - 'a' ; } else { n1 = ch1 - 'A' ; } if (n != n1) { same = false ; } } // If same == true then it means a // substring was found starting at // position therefore all characters // from position to length of substring // found need to be changed therefore // they need to be marked if (same == true ) { for ( int i = position; i < position + k; i++) { pre[i] = true ; } return ; } } // Function performing calculations. public static void solve() { w = z.ToCharArray(); letter = 'd' ; int l = z.Length; int p = letter - 'a' ; for ( int i = 0; i < 100; i++) { pre[i] = false ; } // To verify if any substring is // starting from index i for ( int i = 0; i < l; i++) { for ( int j = 0; j < n; j++) { verify(i, j); } } // Modifying the string w // according to th rules for ( int i = 0; i < l; i++) { if (pre[i] == true ) { if (w[i] == letter) { w[i] = (letter == 'a' ) ? 'b' : 'a' ; } // This condition checks // if w[i]=upper(letter) else if (w[i] == ( char )(( int ) 'A' + p)) { w[i] = (letter == 'a' ) ? 'B' : 'A' ; } // This condition checks if w[i] // is any lowercase letter apart // from letter. If true replace // it with letter else if (w[i] >= 'a' && w[i] <= 'z' ) { w[i] = letter; } // This condition checks if w[i] // is any uppercase letter apart // from letter. If true then // replace it with upper(letter). else if (w[i] >= 'A' && w[i] <= 'Z' ) { w[i] = ( char )(( int ) 'A' + p); } } } Console.WriteLine(w); } // Driver Code public static void Main( string [] args) { n = 3; s[0] = "etr" ; s[1] = "ed" ; s[2] = "ied" ; z = "PEtrUnited" ; solve(); } } // This code is contributed by Shrikanth13 |
输出：
PDddUnitda
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