[树状数组]LightOJ 1085 - All Possible Increasing Subsequences

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.

 

题目大意：找出一个序列中的所有上升序列，很容易想到DP，直接给代码，注释里有说明：

/****  构造n的状态空间， dp[i]表示以i结尾的所有上升序列的个数**  转移为 dp[i] = 1+segma(dp[k])| 0<=k<i && arr[k]<arr[i]**  不用树状数组优化的话转移是O(n)的**  用树状数组优化以后转移就是O(logn)的，需要离散化**/#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#define LL long longusing namespace std;const int MAXN = 100010;int c[MAXN],arr[MAXN],pt[MAXN],t,n,m;LL ans,MOD = 1000000007ll,dp[MAXN];void init(){    memset(c,0,sizeof(c));    ans = 0;}int lowbit(int x){    return x&(-x);}int sum(int pos){    int s = 0;    while(pos>0){        s = (s+c[pos])%MOD;//不取模的话会溢出        pos-=lowbit(pos);    }    return s;}void add(int pos,int val,int n){    while(pos<=n){        c[pos] = (c[pos]+val)%MOD;//不取模的话会溢出        pos+=lowbit(pos);    }}//离散化void disc(){    memcpy(pt,arr,sizeof(arr));    sort(pt,pt+n);    int i;    for(i=1,m=1;i<n;i++){        if(pt[i]!=pt[i-1])pt[m++]=pt[i];    }}int search(int k){    int l=0,r=m-1,m;    while(l<=r){        m = (l+r)>>1;        if(pt[m]==k)return m;        else if(k<pt[m])r=m-1;        else l=m+1;    }    return -1;}int main(){    scanf("%d",&t);    for(int cas=1;cas<=t;cas++){        init();        scanf("%d",&n);        for(int i=0;i<n;i++)scanf("%d",&arr[i]);        disc();        dp[0] = 1;ans = 1;        add(search(arr[0])+1,dp[0],m);        for(int i=1;i<n;i++){            int pos = search(arr[i])+1;            dp[i] = (sum(pos-1)+1)%MOD;            ans = (ans+dp[i])%MOD;            add(pos,dp[i],m);        }        printf("Case %d: %lld\n",cas,ans);    }    return 0;}