本文的目的是实现一个程序，用于计算由一个子字符串重复连接而成的长度为N的二进制字符串的数量。
目标是确定通过重复连接给定文本的单个子字符串，可以创建多少长度为N的二进制字符串，其中N是一个正整数。
实现一个程序，用于计算重复连接子字符串的长度为N的二进制字符串的数量。
Let us take the Input, N = 3
Output: 2
下面列出了长度为N=3的可行二进制字符串，其中重复连接了一个子字符串。
"000":The substring "0" is repeatedly concatenated to form this string. "111":The substring "1" is repeatedly concatenated to form this string.
因此，当我们计算所有这些字符串的总数时，我们得到的和是2。因此，输出为2。
Let us take the Input, N = 8
Output: 16
下面列出了长度为N=8的可行二进制字符串，其中包含了子字符串的重复连接。
“00000000”: The substring "0" is repeatedly concatenated to form this string. “11111111”: The substring "1" is repeatedly concatenated to form this string. “01010101”: The substring "01" is repeatedly concatenated to form this string. “10101010”: The substring "10" is repeatedly concatenated to form this string. "00110011”: The substring "0011" is repeatedly concatenated to form this string. "11001100”: The substring "1100" is repeatedly concatenated to form this string. "11011101”: The substring "1101" is repeatedly concatenated to form this string. "00100010”: The substring "0010" is repeatedly concatenated to form this string. "10111011”: The substring "1011" is repeatedly concatenated to form this string. "01000100”: The substring "0100" is repeatedly concatenated to form this string. "10001000”: The substring "1000" is repeatedly concatenated to form this string. "00010001”: The substring "0001" is repeatedly concatenated to form this string. "11101110”: The substring "1110" is repeatedly concatenated to form this string. "01110111”: The substring "0111" is repeatedly concatenated to form this string. "01100110”: The substring "0110" is repeatedly concatenated to form this string. "10011001”: The substring "1001" is repeatedly concatenated to form this string.
因此，当我们将所有这些字符串的总数相加时，我们得到的和为16。因此，输出结果为16。
为了计算由子串重复连接而成的N长度二进制字符串的数量，我们采用以下方法。
解决这个问题的方法和计算重复连接子字符串的N长度二进制字符串的数量。
可以根据以下事实解决上述问题：每个可行的字符串都包含一个重复的子字符串，该子字符串被连接了C次。因此，所提供的字符串长度N需要被C整除才能生成所有的连续字符串。
因此，首先发现N的所有除数，然后对于每个可能的除数C，发现通过连接它们可以创建的所有潜在字符串的总数；这个数字可以使用2C来确定。为了确定每个递归调用的总计数，对除数C应用相同的技术，然后从2C中减去它。这也将考虑到它们之间存在的重复字符串的数量。
计算下面给定的子字符串重复连接的长度为N的二进制字符串的算法。
第一步 − 开始
第二步 − 定义一个函数来计算长度为N的字符串的数量，使其是其子字符串的连接。
第三步 - 检查状态是否已经计算
第4步 - 存储当前递归调用的结果或计数的值
步骤 5 - 迭代所有除数
第6步 - 返回获得的结果
第7步 − 停止
这是上述算法的C程序实现，用于计算由子字符串重复连接而成的N长度二进制字符串的数量。
// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Storing all the states of recurring recursive
map<int, int> dp;
// Function for counting the number of strings of length n wherein thatstring is a concatenation of its substrings
int countTheStrings(int n){
//the single character cannot be repeated
if (n == 1)
return 0;
// Checking whether the state is calculated already or not
if (dp.find(n) != dp.end())
return dp[n];
// Storing those value of the result or the count for the present recursive call
int res = 0;
// Iterate through all of the divisors
for(int d= 1; d <= sqrt(n); d++){
if (n % d== 0){
res += (1 << d) - countTheStrings(d);
int div1 = n/d;
if (div1 != d and d!= 1)
// Non-Rep = Total - Rep
res += (1 << div1) - countTheStrings(div1);
}
}
// Storing the result of the above calculations
dp[n] = res;
// Returning the obtained result
return res;
}
int main(){
int n = 8;
cout<< "Count of 8-length binary strings that are repeated concatenation of a substring: "<< endl;
cout << countTheStrings(n) << endl;
}
Count of 8-length binary strings that are repeated concatenation of a substring − 16
同样地，我们可以计算出长度为N的二进制字符串，它们是子字符串的重复拼接。
在本文中解决了获取由子字符串重复连接而成的N长度二进制字符串的计数的挑战。
在这里提供了C++编程代码以及计算重复连接子字符串的N长度二进制字符串的算法。
以上是计算长度为N的二进制字符串，它们是子字符串的重复拼接的详细内容。更多信息请关注PHP中文网其他相关文章！
