Problem Link: Light OJ – 1085 – All Possible Increasing Subsequences

Solution Hint: 

This problem can be solved by both BIT and Segment tree. I have applied BIT here. You can read the blog  here to have a better understanding… Here solution hint is given somewhat but you have to do the code.

The most crutial part is how we deal with repeated values !! I request you please do some pen n paper work regarding this n brainstorm yourself that how you will deal with that. If you find a reasonable way, do accordingly, if that works, congratzz!! 😀 if not, no matter, you are probably puzzling things, but one great hint is – there can be many ways to sort it out n it’s easy, believe  me 🙂

Now you can see my code.

I have just subtracted the repeated values elements from my cummulative sum… because, here we are running query for index [i-1] for each i , that is for each element in the vector that is ended woth that element in the resultant subsequence. So of course it’s not a valid subsequence to end with repeated value !!

And the subtraction is less complicated because guess !! I have already sorted the vector on which I am running queries !! So there is no chance of calculating a index which is greater that the next index !! Because the STL sort function automatically sort a vector with pair firstly according to first elements from lower to higher, then if first elements are equal, then by second elements from lower to higher values.

#include <bits/stdc++.h>

#define pf                  printf
#define sf(a)               scanf("%d",&a)
#define sfl(a)              scanf("%lld",&a)
#define sff(a,b)            scanf("%d %d",&a,&b)
#define sffl(a,b)           scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)         scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)        scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d)      scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d)     scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e)   scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e)  scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a)              scanf("%c",&a)
#define pii                 pair<int,int>
#define ms(a,b)             memset(a,b,sizeof(a))
#define pb(a)               push_back(a)
#define pbp(a,b)            push_back({a,b})
#define db                  double
#define ft                  float
#define ll                  long long
#define ull                 unsigned long long
#define pii                 pair<int,int>
#define ff                  first
#define ss                  second
#define sz(x)               x.size()
#define all(x)              x.begin(),x.end()
#define CIN                 ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c)       max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c)       min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define for0(i,n)           for(int i=0;i<n;i++)
#define for1(i,n)           for(int i=1;i<=n;i++)
#define forrev(i,n)         for(int i=n-1; i>=0; i--)
#define forab(i,a,b)        for(int i=a;i<=b;i++)
#define forba(i,b,a)        for(int i=b;i>=a;i--)
#define stlloop(x)          for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b)           __gcd(a, b)
#define lcm(a, b)           ((a)*((b)/gcd(a,b)))
#define case1(z)            cout<<"Case "<<z<<": "
#define case2(z)            printf("Case %d: ",z)
#define PI                  acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty)        tx>=0 && tx<row && ty>=0 && ty<col
#define intlim              2147483648
#define MAX                 200005
#define inf                 100000008

/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/

using namespace std;

ll tree[100005],m=1000000007;
vector<pair<ll,ll> > v;

ll query(ll idx)
{
    ll sum=0;
    while(idx>0)
    {
        sum=((sum%m)+(tree[idx]%m))%m;
        idx-=idx & (-idx);
    }
    return sum;
}

void update(ll idx,ll val,ll n)
{
    while(idx<=n)
    {
        tree[idx]=((tree[idx]%m)+(val%m))%m;
        idx+=idx & (-idx);
    }
}

int main()
{
    //CIN;
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int t;
    sf(t);
    for1(z,t)
    {
        int n;
        sf(n);
        for1(i,n)
        {
            ll x;sfl(x);
            v.pb(make_pair(x,i));
        }
        sort(all(v));
        ll sum,ans=0,x=-inf,k;
        for0(i,sz(v))
        {
            sum=(1+query(v[i].ss-1))%m;
            if(v[i].ff!=x)
            {
                k=sum;
                x=v[i].ff;
            }
            else
            {
                sum-=k;
                k=(k%m+sum%m)%m;
            }
            update(v[i].ss,sum,n);
            ans=(ans%m+sum%m)%m;
        }
        case2(z);
        pf("%lld\n",ans);
        ms(tree,0);
        v.clear();
    }
    return 0;
}