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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

Posted by 200730690105 at 2008-11-20 01:41:25 on Problem 1256
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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