C ++程序在STL中实现Prev_Permutataion
STL中的Prev_permutation用于将范围[first，last]中的元素重新排列到先前的字典上较小的排列。排列是N的每一个！元素可以采取的可能安排。这是一个在STL中实现Prev_permutation的C++程序。
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.
Reverse the array elements.
Do
show(elements) //to display the current content of the array
while (prev_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);
reverse (elements, elements + e);
cout << "The "<<e<<"! possible permutations with ";
cout<<e<<" elements: "<<endl;
do {
show(elements, e);
}
while (prev_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: 10 Enter 4 element: 2 The 4! possible permutations with 4 elements: 10 7 6 2 10 7 2 6 10 6 7 2 10 6 2 7 10 2 7 6 10 2 6 7 7 10 6 2 7 10 2 6 7 6 10 2 7 6 2 10 7 2 10 6 7 2 6 10 6 10 7 2 6 10 2 7 6 7 10 2 6 7 2 10 6 2 10 7 6 2 7 10 2 10 7 6 2 10 6 7 2 7 10 6 2 7 6 10 2 6 10 7 2 6 7 10