next_permutation

// 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]);
}
}