排列

//枚举

#include <cstdio>
int a[100];
int n;

void premutation(int n,int* a,int cur)
{
    if(cur == n)
    {                        
        for(int i = 0; i < n; i++)
        printf("%d",a[i]);
        printf("\n");
        return;
    }
    for(int i  = 0; i < n; i++)
    {
        int ok = 1;
        for(int  j = 0; j < cur; j++)
            if(a[j] == i)             //判断是否当前已选过
                  ok = 0;
        if(ok)
            {a[cur] = i;
        premutation(n,a,cur+1);}                       
    }
}
int main()
{
    scanf("%d",&n);
    premutation(n,a,0);
    return 0;
}

 

//输入数组p，按字典序输出数组a各元素的所有全排列

#include <cstdio>

const int Max = 10000;
int n,a[Max],p[Max];


void permutation(int n,int* p,int* a,int cur)
{
    if(cur == n)
    {
        for(int i = 0; i < n; i++)
            printf("%d",a[i]);
            printf("\n");
    }
    else
        for(int i = 0; i < n; i++)
    {
        int ok = 1;
        for(int j = 0; j < cur; j++)
            if(a[j] == p[i])
              ok = 0;
        if(ok)
        {
            a[cur] = p[i];
            permutation(n,p,a,cur+1);
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i = 0; i < n; i++)
        scanf("%d",&p[i]);
    permutation(n,p,a,0);
    return 0;
}

 

若p中有元素重复，则没有输出，所以需要

//输入数组p，按字典序输出数组a各元素的所有全排列
#include <cstdio>

const int Max = 10000;
int n,a[Max],p[Max];


void permutation(int n,int* p,int* a,int cur)
{
    if(cur == n)
    {
        for(int i = 0; i < n; i++)
            printf("%d",a[i]);
            printf("\n");
    }
    else
        for(int i = 0; i < n; i++)
            if(!i || p[i]!=p[i-1])   //因为p是排好序的，所以只对每个相同元素组做一次递归，并控制输出相同元素个数
    {
        int c1 = 0, c2 = 0;
        for(int j = 0; j < cur; j++)
            if(a[j] == p[i])
               c1 ++;
        for(int j = 0; j < n; j++)
            if(p[j] == p[i])
               c2 ++;
        if(c1 < c2)
        {            //当前a中p[i]比p中的少,控制a中相同元素的个数，若某个数出现n次，最多定位到n-1的时候还能够继续递归
            a[cur] = p[i];
            permutation(n,p,a,cur+1);
        }
    }
}

int main()
{
    scanf("%d",&n);
    for(int i = 0; i < n; i++)
        scanf("%d",&p[i]);
    permutation(n,p,a,0);
    return 0;
}

 

用STL库函数next_permutation直接枚举，适用于可重集

//输入数组p，按字典序输出数组a各元素的所有全排列
#include <cstdio>

const int Max = 10000;
int n,a[Max],p[Max];

#include <algorithm>

using namespace std;

int main()
{
    scanf("%d",&n);
       for(int i = 0; i < n; i++)
        scanf("%d",&p[i]);
        sort(p,p+n);
        do
        {
            for(int i = 0; i < n; i++)
                printf("%d",p[i]);
                printf("\n");
        }while(next_permutation(p,p+n));
        return 0;
}