LightOJ 1085 - All Possible Increasing Subsequences (离散化+树状数组+dp)

1085 - All Possible Increasing Subsequences

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Time Limit: 3 second(s)

Memory Limit: 64 MB

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.

Sample Input	Output for Sample Input
3 3 1 1 2 5 1 2 1000 1000 1001 3 1 10 11	Case 1: 5 Case 2: 23 Case 3: 7

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 }