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;
}
Fall out, Codes! 