1085 - All Possible Increasing Subsequences[树状数组]

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.      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.

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PROBLEM SETTER: JANE ALAM JAN

题目大意：给你n个数，问你其中的单调上升子区间的个数

可以使用dp的想法

我们假设dp[i]代表：以数字i结尾的单调上升子区间的个数

那么结尾数字为i的单调上升子区间就是所有结尾数字小于i的单调上升子区间后面加上一个i，然后就是一个i也可以单独成为一个单调上升子区间

那么dp[i] = dp[1]+dp[2]+...+dp[i-1]+1

注意上式有前缀和

因此我们可以先把数组进行离散化，然后用树状数组进行维护

代码如下：

#include <bits/stdc++.h>using namespace std;const int MAX_N = 1e5+100;const int MOD = 1000000007;int bit[MAX_N],n,T,a[MAX_N],b[MAX_N];void add(int i,int x){    while(i<=n){        bit[i] += x;        while(bit[i]>MOD) bit[i] -= MOD;        i+=i&-i;    }}int sum(int i){    int res = 0;    while(i>0){        res += bit[i];        i -= i&-i;        while(res>MOD) res -= MOD;    }    return res;}int main(){    scanf("%d",&T);    for(int t=1;t<=T;t++){        memset(bit,0,sizeof(bit));        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        scanf("%d",&n);        for(int i=0;i<n;i++){            scanf("%d",&a[i]);            b[i] = a[i];        }                // 以下代码为离散化操作        sort(a,a+n);        int N = unique(a,a+n)-a;        for(int i=0;i<n;i++){            b[i] = lower_bound(a,a+N,b[i])-a;            b[i]+=1;        }                // 树状数组维护值        for(int i=0;i<n;i++){            int s = sum(b[i]-1);            s++;            add(b[i],s);        }        int ans = sum(n);        printf("Case %d: %d\n",t,ans);    }    return 0;}