1085 - All Possible Increasing Subsequences

An increasing subsequence from a sequence A₁, A₂ ... A_n is defined by A_i1, A_i2 ... A_ik, where the following properties hold

1. i₁ < i₂ < i₃ < ... < i_k and
2. 2.      A_i1 < A_i2 < A_i3 < ... < A_ik

Now you are given a sequence, you have to find the number of all possible increasing subsequences.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 10⁵) denoting the number of elements in the initial sequence. The next line will contain n integers separated by spaces, denoting the elements of the sequence. Each of these integers will be fit into a 32 bit signed integer.

Output

For each case of input, print the case number and the number of possible increasing subsequences modulo 1000000007.

Notes

1. For the first case, the increasing subsequences are (1), (1, 2), (1), (1, 2), 2.
2. Dataset is huge, use faster I/O methods.

题目大意：

　　就是说，给你一个长度为n的序列，然后，让你求出这个序列中所有的上升子序列的个数

解题思路：

　　直接把tree[i]当做dp[i]来用，表示的是以a[i]结尾的上升序列的最大个数

代码：

 1 # include<iostream>
 2 # include<cstdio>
 3 # include<algorithm>
 4 # include<cstring>
 5
 6 using namespace std;
 7
 8 # define MAX 500000+4
 9 # define MOD 1000000007
10
11 typedef long long LL;
12
13 LL a[MAX];
14 LL data[MAX];
15 int tree[MAX];
16 int n,cc;
17
18 void discrete()
19 {
20     memset(data,0,sizeof(data));
21     for ( int i = 0;i < n;i++ )
22     {
23         data[i] = a[i];
24     }
25     sort(data,data+n);
26     cc = unique(data,data+n)-data;
27     for ( int i = 0;i < n;i++ )
28     {
29         a[i] = 1+lower_bound(data,data+cc,a[i])-data;
30     }
31 }
32
33
34 void update ( int pos,int val )
35 {
36     while ( pos <= n )
37     {
38         tree[pos]=(tree[pos]+val)%MOD;
39         pos += pos&(-pos);
40     }
41 }
42
43 LL read ( int pos )
44 {
45     LL sum = 0;
46     while ( pos>0 )
47     {
48         sum=(sum+tree[pos])%MOD;
49         pos-=pos&(-pos);
50     }
51     return sum%MOD;
52 }
53
54
55 int main(void)
56 {
57     int icase = 1;
58     int t;cin>>t;
59     while ( t-- )
60     {
61         while ( cin>>n )
62         {
63         if ( n==0 )
64             break;
65         LL ans = 0;
66         for ( int i = 0;i < n;i++ )
67         {
68             cin>>a[i];
69         }
70         discrete();
71         memset(tree,0,sizeof(tree));
72         for ( int i = 0;i < n;i++ )
73         {
74             update(a[i],read(a[i]-1)+1);
75         }
76         printf("Case %d: ",icase++);
77         cout<<read(cc)%MOD<<endl;
78         }
79     }
80
81     return 0;
82 }