include

using namespace std;

define ll long long

define no cout<<"No"<

define yes cout<<"Yes"<

define mod 1000000007

define nt 10 define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)

void solve() { vectorv; sets; mapmp; ll n,m,j,i,y,z,l,r,d,x,k; cin>>n>>k; ll a[n+1]; ll c=0; for(i=0;i>a[i]; if(a[i]>k)c++; } x=0; ll mx=-1,ans=0; for(i=0;ik) { ans+=(i-mx)*(n-i); mx=i; } } cout<

} int main() { fast; int T=1; cin>>T; while(T--)solve(); return 0; }

2 years ago+ 0 comments

Hint: prefix array

3 years ago+ 0 comments

Just a humble request to all others who have solved questions like these, please refrain from showing the solutions or giving links to it here in the discussion as it destroys the purpose of solving such problems. Please try to give only intuitions to guide those who need assistance rather than giving them the code.

Thank you, Sincerely, akshay_123

4 years ago+ 0 comments

ruby:

gets.to_i.times do
    n,k = gets.strip.split.map &:to_i
    a = gets.strip.split.map &:to_i
    max, prev = (1..n).reduce(:+), 0
    for i in 0...n do
        if a[i] > k
            max -= (1..prev).reduce(:+) || 0
            prev = 0
        else
            prev +=1 end
        i += 1 end
    max -= (1..prev).reduce(:+) || 0
    puts max end

5 years ago+ 0 comments

for a0 in range(int(input())):
    x=[int(y)for y in input().split()]
    n,k=x[0],x[1]
    ar=[int(y)for y in input().split()]
    s=n
    b=[]
    for i in  range(n):
        if ar[i]>k:
            b.append(s)
        s-=1
    first=0
    if len(b)==0:
        print(0)
        continue
    total=0
    for i in range(n):
        if ar[i]<=k:
            total+=b[first]
        elif ar[i]>k:
            total+=b[first]
            first+=1
            if first==len(b):
                break
    print(total)

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