重新排列给定的字符串，使所有素数多个索引都具有相同的字符

给定一个大小为N的字符串str 。任务是找出是否可以重新排列字符str 中的字符串，以便对于任何素数 p <= N 和从 1 到 N/p 的任何整数 i 都必须满足条件 str _p = str _p*i .如果不可能进行任何此类重新排列，则打印 -1。

例子：

Input : str = “aabaaaa”
Output : baaaaaa
Size of the string is 7.
Indexes 2, 4, 6 contains same character.
Indexes 3, 6 contains the same character.

Input : str = “abcd”
Output : -1

编程需要懂一点英语

方法：

除了第一个位置和数量为大于 N/2 的素数的位置之外的所有位置必须具有相同的符号。
剩余位置可以有任何符号。这个位置一直使用筛子。
如果字符串中出现次数最多的元素小于这些位置，则打印 -1。

下面是上述方法的实现：

C++

// CPP program to rearrange the given 
// string such that all prime multiple
// indexes have same character
#include 
using namespace std;
#define N 100005
  
// To store answer
char ans[N];
int sieve[N];
  
// Function to rearrange the given string
// such that all prime multiple indexes
// have the same character.
void Rearrange(string s, int n)
{
    // Initially assume that we can kept
    // any symbol at any positions.
    // If at any index contains one then it is not
    // counted in our required positions
    fill(sieve + 1, sieve + n + 1, 1);
  
    // To store number of positions required
    // to store elements of same kind
    int sz = 0;
  
    // Start sieve
    for (int i = 2; i <= n / 2; i++) {
        if (sieve[i]) {
            // For all multiples of i
            for (int j = 1; i * j <= n; j++) {
                if (sieve[i * j])
                    sz++;
                sieve[i * j] = 0;
            }
        }
    }
  
    // map to store frequency of each character
    map m;
    for (auto it : s)
        m[it]++;
  
    // Store all characters in the vector and
    // sort the vector to find the character with
    // highest frequency
    vector > v;
    for (auto it : m)
        v.push_back({ it.second, it.first });
    sort(v.begin(), v.end());
  
    // If most occured character is less than
    // required positions
    if (v.back().first < sz) {
        cout << -1;
        return;
    }
  
    // In all required positions keep
    // character which occured most times
    for (int i = 2; i <= n; i++) {
        if (!sieve[i]) {
  
            ans[i] = v.back().second;
        }
    }
  
    // Fill all other indexes with
    // remaining characters
    int idx = 0;
    for (int i = 1; i <= n; i++) {
        if (sieve[i]) {
            ans[i] = v[idx].second;
            v[idx].first--;
            // If character frequency becomes
            // zero then go to next character
            if (v[idx].first == 0)
                idx++;
        }
        cout << ans[i];
    }
}
  
// Driver code
int main()
{
    string str = "aabaaaa";
  
    int n = str.size();
  
    // Function call
    Rearrange(str, n);
  
    return 0;
}

Python3

# Python3 program to rearrange the given 
# string such that all prime multiple 
# indexes have same character 
  
N = 100005
  
# To store answer 
ans = [0]*N; 
# sieve = [1]*N; 
  
# Function to rearrange the given string 
# such that all prime multiple indexes 
# have the same character. 
def Rearrange(s, n) : 
  
    # Initially assume that we can kept 
    # any symbol at any positions. 
    # If at any index contains one then it is not 
    # counted in our required positions 
    sieve = [1]*(N+1);
      
    # To store number of positions required 
    # to store elements of same kind
    sz = 0;
      
    # Start sieve
    for i in range(2, n//2 + 1) :
        if (sieve[i]) :
              
            # For all multiples of i
            for j in range(1, n//i + 1) :
                if (sieve[i * j]) :
                    sz += 1;
                sieve[i * j] = 0;
                  
    # map to store frequency of each character 
    m = dict.fromkeys(s,0);
      
    for it in s :
        m[it] += 1;
          
    # Store all characters in the vector and 
    # sort the vector to find the character with 
    # highest frequency
    v = [];
    for key,value in m.items() :
        v.append([ value, key] );
      
    v.sort();
      
    # If most occured character is less than
    # required positions
    if (v[-1][0] < sz) :
        print(-1,end="");
        return;
          
    # In all required positions keep
    # character which occured most times
    for i in range(2, n + 1) :
        if (not sieve[i]) :
            ans[i] = v[-1][1];
              
    # Fill all other indexes with 
    # remaining characters 
    idx = 0;
      
    for i in range(1, n + 1) :
        if (sieve[i]):
            ans[i] = v[idx][1];
            v[idx][0] -= 1;
              
            # If character frequency becomes 
            # zero then go to next character
            if (v[idx][0] == 0) :
                idx += 1;
                  
        print(ans[i],end= ""); 
          
          
# Driver code 
if __name__ == "__main__" : 
  
    string = "aabaaaa"; 
  
    n = len(string); 
  
    # Function call 
    Rearrange(string, n); 
  
# This code is contributed by AnkitRai01

输出：

baaaaaa