1.
There is a function next_permutation(start,end) in the algorithm library of template C++. The role is to find the next permutation of a sort, you can traverse all permutations. Note that if it does not end, the loop will continue, and the result will appear the same arrangement. The opposite function is prev_permutation(start,end), which is a function to find the previous permutation of a sort.
For the next_permutation function, its function prototype is:
include
bool next_permutation(iterator start, iterator end)
// When there is no next permutation in the current sequence, the function returns false, otherwise it returns true.
#include<bits/stdc++.h>
using namespace std;
int n; // n表示序列中数的个数
int a[10005];
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
cin >> a[i];
do
{
for (int i = 0; i < n; i++) // 打印
cout << a[i] << " ";
cout << endl;
}while (next_permutation(a, a + n));
return 0;
}
It should be emphasized that next_permutation() needs to sort the array to be arranged in ascending order before using it, otherwise it can only find out the total number of permutations after the sequence.
A sample question corresponding to the non-template //z code
#include<bits/stdc++.h>
int cnt=0;
using namespace std;
typedef long long ll;
const int maxn=1e6+199;
int len1,len2,len3;
ll a[maxn];
int k,n,m;
void dop()
{
for (k = n - 1; k >= 1; k--)
{
if (a[k] < a[k + 1]) //找到最后可增加的位，即定位
break;
}
for (int i = n; i > k; i--)
{
if (a[i] > a[k])
{
swap(a[i], a[k]); //找到最小可增加的数字 交换
break;
}
}
sort(a + k + 1, a + n + 1); //对后面的升序排列
}
int main(){
cin>>n;
cin>>m;
for(int i=1;i<=n;i++)
cin>>a[i];
int i=0;
while(i<m)
{
i++;
dop();
}
for(int i=1;i<=n;i++)
{
if(i==1)
cout<<a[i];
else
cout<<' '<<a[i];
}
return 0;
}