设计递归算法生成n个元素的所有排列对象

Question

如题，C/C++，在线等

Answer 1

#include<iostream>
#include<cmath>
using namespace std;

int count(int n)

//算n的阶乘——因为n个数能组成n!个数
{

if(n<1)

{

cout<<"输入也错！";

exit(0);

}

else

if(n==1)

return 1;

else

return count(n-1)*n;
}

int pow10(int n)
//算出10的n次方
{

int a;

if(n==0)

return 1;

else

for(int i=1;i<=n;i++)

a=10*pow10(n-1);

return a;
}
int * comm(int n)

//组合n!个数（这里用递归算）
{

int *a=new int[count(n)];

if(count(n)==1)

a[0]=1;

else

{

int *b=new int[count(n-1)];

b=comm(n-1);

for(int i=0;i<count(n-1);i++)

for(int j=0;j<n;j++)

a[i*n+j]=(b/pow10(j)*10+n)*pow10(j)+b%pow10(j);

}

return a;
}
void main()
{

int n;

cout<<"请输入n=";

cin>>n;

int *a=new int[count(n)];

a=comm(n);

cout<<"1-"<<n<<"自然数所有的排列组合为：\
";

for(int i=0;i<count(n);i++)

cout<<a<<" ";
}

=======================================
#define MAX 1000
#include<stdio.h>

void DispArrangement(int a[MAX], int n, int deepth)
{

int i, temp;

if(deepth == 1) {

for(i = 1; i <= n; i ++) {

printf("%d", a);

}

printf("\
");

} else {

for(i = 1; i <= deepth; i ++ ){

temp = a[n - deepth + 1];

a[n - deepth + 1] = a[n - deepth + i];

a[n - deepth + i] = temp;

DispArrangement(a, n, deepth - 1);

temp = a[n - deepth + 1];

a[n - deepth + 1] = a[n - deepth + i];

a[n - deepth + i] = temp;

}

}
}

int main(void)
{

int i, n, a[MAX];

scanf("%d", &n);

for(i = 1; i <= n; i ++ ) {

a = i;

}

DispArrangement(a, n, n);

return 0;
}

Answer 2

//输出其中包含n个数的所有排列
#include <iostream>
using namespace std;

int data[100];
void DPpl(int num,int m,int n,int depth)
{
if(depth==n)
{
for(int i=0;i<n;i++)
cout<<data[i]<<" ";
cout<<endl;
}
for(int j=0;j<m;j++)
{
if((num&(1<<j))==0)
{
data[depth]=j+1;
DPpl(num+(1<<j),m,n,depth+1);
}
}
}
int main()
{
//DPpl(0,5,1,0);
//DPpl(0,5,2,0);
DPpl(0,5,3,0);//5个数输出包含其中3个数的排列
//DPpl(0,5,4,0);
//DPpl(0,5,5,0);
return 0;
}

Answer 3

12523你啊

Answer 4

#include<stdio.h>
#include<stdlib.h>

struct cc{
char c;
int s;
};

void f(cc* sc, char* t, int n, int l)
{
int i;
for(i=0;i<n;i++){
if(sc[i].s == 1){
sc[i].s = 0;
t[l] = sc[i].c;
if(l<n-1)
f(sc, t, n, l+1);
sc[i].s = 1;
}
}
if(l == n-1){
t[n]='\0';
printf("%s\n", t);
}
}
int main()
{
int n,i;
scanf("%d", &n);
char* arr = (char*)malloc(sizeof(char)*n);
scanf("%s", arr);
cc* ac = (cc*)malloc(sizeof(cc)*n);
for(i = 0 ;i<n;i++){
cc c;
c.c = arr[i];
c.s = 1;
ac[i] = c;
}
char* temp = (char*)malloc(sizeof(char)*(n+1));
f(ac, temp, n, 0);
return 0;
}本回答被网友采纳