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用普通的next_permutation(a,a+n)也能过,只要对字符串进行处理比如说开一个int数组,把'A'转为1,'a'转为2,'B'转为3...
总之大写转偶数序列,小写转奇数(一一对应)
每next_permutation(a,a+n)一次就把int转成字符串输出(还是那个法则,只不过逆了过来)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
char a[15];
int b[15];
int cmp ( const void *a , const void *b )
{
return *(int *)a - *(int *)b;
}
void change1()
{
int n,i;
n=strlen(a);
for(i=0;i<=n-1;i++)
{
if(a[i]>='A' && a[i]<='Z')
{
b[i]=2*(a[i]-'A')+1;
}
else if(a[i]>='a' && a[i]<='z')
{
b[i]=2*(a[i]-'a')+2;
}
}
}
void change2(int n)
{
int i;
for(i=0;i<=n-1;i++)
{
if(b[i]%2==1)
{
a[i]=(b[i]-1)/2+'A';
}
else if(b[i]%2==0)
{
a[i]=(b[i]-2)/2+'a';
}
}
}
int main()
{
int t;
int n;
scanf("%d",&t);
while(t--)
{
scanf("%s",a);
n=strlen(a);
change1();
qsort(b,n,sizeof(b[0]),cmp);
do
{
change2(n);
printf("%s\n",a);
}while(next_permutation(b,b+n));
}
return 0;
}
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