Evaluate the full arrangement of numbers or strings

Taking numbers as an example: there is an array of numbers A: 1 2 3 4, its full row by dictionary sequence column:

1 2 3 4
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 4 1
..........

N in total! Now input a number K and output its k-th Arrangement

Method: The idea of using Kanto code is actually to find the number at each position: the number at the first position, the number at the second position ....

Solution: calculate the number at each position in order. Here K = 8 is assumed; array: 1 2 3 4; Length: 4; position P indicates the number of P in the array: P = K/(n-1 )!
1. Calculate the number at the first position. P indicates the position of the first number in array.

1) P = 8/(4-1 )! = 8/6 = 1; ----------> the first number is a [1] = 2 2) Remove a [1] From array A, and the remaining number is 1 3 4 3) k = K % (4-1 )! = 8% 6 = 2

2. Repeat the process in step 1 to find the number at the second position:

1) At this time K = 2, array a becomes: 1 3 4, array length is 3 2) p = 1/(3-1 )! = 2/2 = 1; ----------> the second number is a [1] = 3 3) Remove a [1] From array A, and the remaining number is 1 4 4 4) k = K % (3-1 )! = 2% 2 = 0

3. Calculate the number at the 100th position and repeat the process of Step 1:

1) k = 0 at this time, and the loop ends. Before the end, we need to determine the number of remaining positions! In fact, it is very simple: no operation is required, that is, the remaining elements in array a are output in reverse order: At this time, the remaining numbers in array a are: 1, 4; so the third number in the result sequence is 4, number 4: 1

The condition for loop end is k = 0.

Final result: the 8th sequences are: 2 3 4 1

Code:

#include<iostream>#include<vector>#include<algorithm>using namespace std;int getNjiecheng(int n){     int res=1;     for(int i=1;i<=n;i++){          res*=i;     }     return res;}vector<int> getPermutationSequence(int k,vector<int> v){     vector<int> res;     int n=v.size();     if(k>getNjiecheng(n))return res;     while(k>0){          int j=getNjiecheng(n-1);          int p = k/j;          k=k%j;          if(k==0){               p-=1;               res.push_back(v[p]);               vector<int>::iterator iter=v.begin()+p;               v.erase(iter);               reverse(v.begin(),v.end());               for(int i=0;i<v.size();i++) res.push_back(v[i]);               return res;          }          else{                res.push_back(v[p]);               vector<int>::iterator iter=v.begin()+p;               v.erase(iter);               n--;          }     }}int main(){     vector<int> v;     for(int i=1;i<=4;i++)          v.push_back(i);     vector<int> res=getPermutationSequence(24,v);     for(vector<int>::iterator iter=res.begin();iter!=res.end();iter++){          cout<<*iter<<"  ";     }     cout<<endl;     return 0;}

If we want n! In full order: Just loop n! Call getpermutationsequence (25, V );

for(int i=1;i<=N!;i++){     getPermutationSequence(i,v);}

Evaluate the full arrangement of numbers or strings