原题地址

法一：dfs，可直接搜索，数据规模小或需要输出具体排列时使用

cpp
#include<bits/stdc++.h>
using namespace std;

#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); ++i)
#define rrep(i, a, b) for(int i = (int)(a); i < (int)(b); ++i)
#define lep(i, a, b) for(int i = (int)(a); i >= (int)(b); i--)
#define UPS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define endn puts("")

typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double EPS = 1e-7, PI = acos(-1);
/*=======================================*/
int book[10]= {0},s=0;
int n,arr[10]= {0};

void dfs(){
	if(s==n){
		rrep(i,0,n){
			printf("%5d",arr[i]);
		}
		endn;
	}
	rep(i,1,n){
		if(!book[i]){
			book[i]++;
			arr[s++]=i;
			dfs();
			s--;
			book[i]--;
		}
	}
}

int main() {
	UPS;
	cin>>n;
	dfs();
	return 0;
}

法二：使用c++自带的库函数next_permutation()，此库函数会将一段数组变为它的下一个全排列，只需统计次数，然后调用即可。

cpp
#include<bits/stdc++.h>
using namespace std;

#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); ++i)
#define rrep(i, a, b) for(int i = (int)(a); i < (int)(b); ++i)
#define lep(i, a, b) for(int i = (int)(a); i >= (int)(b); i--)
#define UPS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define endn puts("")

typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double EPS = 1e-7, PI = acos(-1);
/*=======================================*/
int sum=1;
int n,arr[10]= {0};

int main() {
	UPS;
	cin>>n;
	rrep(i,0,n) arr[i]=n-i,sum*=arr[i];
	for(int i=1;i<=sum;i++){
		next_permutation(arr,arr+n);
		rrep(i,0,n) {
			printf("%5d",arr[i]);
		}
		endn;
	}
	return 0;
}