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：

# include<iostream>
# include<cstdio>
# include<algorithm>
# include<cstring>

using namespace std;

# define MAX 500000+4
# define MOD 1000000007

typedef long long LL;

LL a[MAX];
LL data[MAX];
int tree[MAX];
int n,cc;

void discrete()
{
    memset(data,0,sizeof(data));
    for ( int i = 0;i < n;i++ )
    {
        data[i] = a[i];
    }
    sort(data,data+n);
    cc = unique(data,data+n)-data;
    for ( int i = 0;i < n;i++ )
    {
        a[i] = 1+lower_bound(data,data+cc,a[i])-data;
    }
}


void update ( int pos,int val )
{
    while ( pos <= n )
    {
        tree[pos]=(tree[pos]+val)%MOD;
        pos += pos&(-pos);
    }
}

LL read ( int pos )
{
    LL sum = 0;
    while ( pos>0 )
    {
        sum=(sum+tree[pos])%MOD;
        pos-=pos&(-pos);
    }
    return sum%MOD;
}


int main(void)
{
    int icase = 1;
    int t;cin>>t;
    while ( t-- )
    {
        while ( cin>>n )
        {
        if ( n==0 )
            break;
        LL ans = 0;
        for ( int i = 0;i < n;i++ )
        {
            cin>>a[i];
        }
        discrete();
        memset(tree,0,sizeof(tree));
        for ( int i = 0;i < n;i++ )
        {
            update(a[i],read(a[i]-1)+1);
        }
        printf("Case %d: ",icase++);
        cout<<read(cc)%MOD<<endl;
        }
    }

    return 0;
}