next_permutation

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摘自http://old.blog.edu.cn/user4/248323/archives/2007/1935675.shtml

STL就是好啊，next_permutation可以计算一组数据的全排列

// next_permutation＃i nclude <iostream>＃i nclude <algorithm>using namespace std;int main () {  int myints[] = {1,2,3};  cout << "The 3! possible permutations with 3 elements:/n";  sort (myints,myints+3);  do {    cout << myints[0] << " " << myints[1] << " " << myints[2] << endl;  } while ( next_permutation (myints,myints+3) );  return 0;}

这样就可以排除n个数的全排列了

int main(){
int a[] = {3,1,2};
do{
     cout << a[0] << " " << a[1] << " " << a[2] << endl;
}
 while (next_permutation(a,a+3));
 return 0;
}

如果是这样的话，因为原序列没有排序，就从当前序列开始，只输出比它大大序列，若已经到最大序列，返回值为false，序列变成最小的序列，就不再执行了

prev_permutation与next_permutation相反，由原排列得到字典序中上一次最近排列

http://support.microsoft.com/kb/157869/zh-tw

http://blog.bc-cn.net/user20/130988/archives/2007/6553.shtml中有详细的解析


//学习next_permutation函数 1833的代码
＃i nclude<iostream>
＃i nclude<algorithm>
using namespace std;
int main()
{
    int ncase,n,k,i;
    int s[1024];
    scanf("%d",&ncase);
    while(ncase--)
    {
       scanf("%d%d",&n,&k);
       for(i=0;i<n;i++)
           scanf("%d",&s[i]);
       for(i=0;i<k;i++)
           next_permutation(s,s+n);
       for(i=0;i<n-1;i++)
          printf("%d ",s[i]);
       printf("%d/n",s[i]);      
    }
}