hdu 3998 Sequence

There is a sequence X (i.e. x[1], x[2], ..., x[n]). We define increasing subsequence of X
as x[i1], x[i2],...,x[ik], which satisfies follow conditions:

1) x[i1] < x[i2],...,<x[ik];

2) 1<=i1 < i2,...,<ik<=n

As an excellent program designer, you must know how to find the maximum length of the

increasing sequense, which is defined as s. Now, the next question is how many increasing

subsequence with s-length can you find out from the sequence X.

For example, in one case, if s = 3, and you can find out 2 such subsequence A and B from X.

1) A = a1, a2, a3. B = b1, b2, b3.

2)
Each ai or bj(i,j = 1,2,3) can only be chose once at most.

Now, the question is:

1) Find the maximum length of increasing subsequence of X(i.e. s).

2) Find the number of increasing subsequence with s-length under conditions described (i.e. num).

InputThe input file have many cases. Each case will give a integer number n.The next line will

have n numbers.OutputThe output have two line. The first line is s and second line is num.Sample Input

4
3 6 2 5

Sample Output

2
2
 
最长上升子序列。
代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#define Max 1000
using namespace std;
int n,s[Max],vis[Max];///vis标记是否已经使用
int maxl()
{
    int res = ,t[Max];
    for(int i = ;i < n;i ++)
    {
        if(vis[i])continue;
        if(!res || t[res - ] < s[i])
        {
            t[res ++] = s[i];
            vis[i] = ;
        }
        else
        {
//            *lower_bound(t,t + res,s[i]) = s[i];
            int l = ,r = res,mid;
            while(l < r)
            {
                mid = (l + r) / ;
                if(t[mid] >= s[i])r = mid;
                else l = mid + ;
            }
            t[l] = s[i];
        }
    }
    return res;
}
int main()
{
    while(scanf("%d",&n) != EOF)
    {
        for(int i = ;i < n;i ++)
        {
            scanf("%d",&s[i]);
        }
        memset(vis,,sizeof(vis));
        int m = maxl(),c = ;
        while(maxl() == m)c ++;
        printf("%d\n%d\n",m,c);
    }
}