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C ++程序在STL中实现Next_Permutation

2024-04-20 21:45:03 359

STL中的next_permutation用于将[first，last]范围内的元素重新排列到下一个字典上更大的排列。排列是N的每一个！元素可以采取的可能安排。这是C++程序，用于在STL中实现Next_permutation。

算法

Begin
   Define one integer array variable elements[].
   Get the number of data e from the user.
   Initialize the array elements[] with e number of data from the keyboard.
   Sort all the array elements.
   do
      show(elements, e) //to display the current content of the array
   while (next_permutation(elements, elements + e))
End

范例程式码

#include<iostream>
#include <algorithm>
using namespace std;
void show(int a[], int n) {
   for(int i = 0; i < n; i++) {
      cout<<a[i]<<" ";
   }
   cout<<endl;
}
int main () {
   int e, i;
   cout<<"Enter number of elements to be inserted: ";
   cin>>e;
   int elements[e];
   for (i = 0; i < e; i++) {
      cout<<"Enter "<<i + 1<<" element: ";
      cin>>elements[i];
   }
   sort (elements, elements + e);
   cout << "The "<<e<<"! possible permutations with ";
   cout<<e<<" elements: "<<endl;
   do {
      show(elements, e);
   }
   while (next_permutation(elements, elements + e));
   return 0;
}

输出结果

Enter number of elements to be inserted: 4
Enter 1 element: 7
Enter 2 element: 6
Enter 3 element: 2
Enter 4 element: 10
The 4! possible permutations with 4 elements:
2 6 7 10
2 6 10 7
2 7 6 10
2 7 10 6
2 10 6 7
2 10 7 6
6 2 7 10
6 2 10 7
6 7 2 10
6 7 10 2
6 10 2 7
6 10 7 2
7 2 6 10
7 2 10 6
7 6 2 10
7 6 10 2
7 10 2 6
7 10 6 2
10 2 6 7
10 2 7 6
10 6 2 7
10 6 7 2
10 7 2 6
10 7 6 2

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C ++程序在STL中实现Next_Permutation

算法

范例程式码

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