comments about the problem at whu.edu.cn:

the answer is C(2n, n)/(n+1). (you may have to change n to n+1 or n-1).

Reason: for n nodes, you can have C(2n, n)/(n+1) different binary search trees.

C(2n, n)/(n+1) is called the Catlan number, if I did not remember wrong.

This problem is esentially the same as #1522 at nku.edu.cn. Following is my submission at nku:

main(){int n,i,n2,c=0;long long r;while(scanf(\"%d\",&n)!=-1){++c;r=1;n2=(n<<1);for(i=1;i<=n;++i){r*=(n2-i+1);r/=i;}r/=(n+1);printf(\"Case %d:\n\",c);printf(\"%lld\n\",r);}}

LS再想想那个搜索题目

[此贴子已经被作者于2007-9-16 13:42:02编辑过]

i would think that search problem needs some graph algorithm --- which I am not familar with so that I did not even try it.

HJin

逆序数那个你到PKU上去提交..注意那个64位就可以了