Lightoj 1085-all Possible Increasing subsequences tree array + discrete

http://www.lightoj.com/volume_showproblem.php?problem=1085

Test Instructions: the number of ascending subsequence sequences for a sequence.

idea: to find the law can be found that a number as the number of the end of the type of the number of all smaller than its case +1. Use a tree-like array to find it. C++11 Auto is not available on Lightoj/.\

/** @Date: 2016-12-01-21.58 * @Author: Lweleth ([email protected]) * @Link: https://github.com/* @Versi On: */#include <bits/stdc++.h> #define LL long long#define PII pair<int,int> #define MP (x, y) Make_pair ((x), (Y )) #define FI first#define se second#define PB (x) push_back ((x)) #define MMG (x) memset ((x), -1,sizeof (x)) #define MMF (x) mem  Set ((x), 0,sizeof (x)) #define MMI (x) memset ((x), INF, sizeof (x)) using namespace Std;const int INF = 0x3f3f3f3f;const int N =        1e5+20;const int mod = 1000000007;int C[n], a[n];void Add (int p, int v) {while (P < N) {C[p] + = V;        C[P]%= mod;    p + = (-P) & p;    }}int sum (int p) {int ans = 0;        while (p) {ans + = c[p];        Ans%= MoD;    P-= (-P) & p; } return ans;    int main () {int T;    int cnt = 0;    Cin >> T;        while (t--) {MMF (C);        int n;        scanf ("%d", &n);        Map<int, int>q;   for (int i = 1; I <= n; i++) {         scanf ("%d", &a[i]);        Q[a[i]] = 1;        } int c = 1;        for (Auto i = Q.begin (); I! = Q.end (); i++) I->se = c + +;        /*map<int, int>::iterator it; for (it = Q.begin (); It! = Q.end (); it++) It->se = c++;*/for (int i = 1; I <= n; i++)//Find the law to find a number            As the number of species of the end number for all cases smaller than its number +1 {int x = q[a[i]];        Add (x, sum (x-1) + 1);    } printf ("Case%d:%d\n", ++cnt, SUM (c-1)); } return 0;}

Lightoj 1085-all Possible increasing subsequences tree array + discrete