lightoj1085 - All Possible Increasing Subsequences（树状数组）

题意：给你一个数列，问你有几个递增子序列（单个数字也算一个）

a1, a2, a3....an以ai结尾的递增子序列数量为在此之前出现的并且ak(k<i&&ak<ai)比他小的ak结尾的递增子序列数量之和+1(它本身)

这样就可以用树状数组做了，不断插入并且计算在这之前比他小的子序列和.

因为ai可能很大，所以此题需要离散化一下。

还有别忘了取模。

代码：

#include<bits/stdc++.h>using namespace std;const int maxn = 1e5+5;const int mod = 1e9+7;int tree[maxn], b[maxn];struct node{    int index, val;    bool operator < (const node &a) const    {        return val < a.val;    }}a[maxn];int lowbit(int x) { return x&(-x); }void update(int pos, int val){    while(pos < maxn)    {        tree[pos] = (tree[pos]+val)%mod;        pos += lowbit(pos);    }}int query(int pos){    int sum = 0;    while(pos)    {        sum = (sum+tree[pos])%mod;        pos -= lowbit(pos);    }    return sum;}int main(void){    int n, t, ca = 1;    cin >> t;    while(t--)    {        memset(tree, 0, sizeof(tree));        scanf("%d", &n);        for(int i = 1; i <= n; i++)            scanf("%d", &a[i].val), a[i].index = i;        sort(a+1, a+1+n);        for(int i = 1; i <= n; i++)        {            b[a[i].index] = i;            if(a[i].val == a[i-1].val)                b[a[i].index] = b[a[i-1].index];        }        int temp, sum = 0;        for(int i =  1; i <= n; i++)        {            temp = (query(b[i]-1)+1)%mod;            sum = (sum+temp)%mod;            update(b[i], temp);        }        printf("Case %d: %d\n", ca++, sum);    }    return 0;}

1085 - All Possible Increasing Subsequences

   PDF (English)StatisticsForum

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.