Codeforces Round #725 (Div. 3) 1538C. Number of Pairs Solution

Explain: 

Here given condition is      l<=a[i]+a[j]<=r  

From the condition we get    a[j]<=r-a[i]

                                             l-a[i]<=a[j]

 

              from two equation we can write     l-a[i]<=a[j]<=r-a[i];

from the equation we can find the upperbound(r-a[i])-lowerbound(l-a[i])  count for all index.

ai+aj≤r l≤ai+aj≤r l

Problem Link:  https://codeforces.com/problemset/problem/1538/C

Solve in C++:  

///La ilaha illellahu muhammadur rasulullah

///******Bismillahir-Rahmanir-Rahim******///

///Abul Hasnat  Tonmoy

///Department of CSE,23rd batch

///Islamic University,Bangladesh

///**********ALLAH IS ALMIGHTY************///

#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long  t,n,i,j,l,r,a,x,y,b,cn,u,v;

    cin>>t;

    while(t--)

    {

        cin>>n>>l>>r;

        int a[n+9];

        cn=0;

        for(i=0; i<n; i++)cin>>a[i];

        sort(a,a+n);

        for(i=0; i<n; i++)

        {

            u=l-a[i];

            v=r-a[i];

            x=lower_bound(a+i+1,a+n,u)-a;

            y=upper_bound(a+i+1,a+n,v)-a;

            cn+=abs(y-x);

        }

        cout<<cn<<endl;

    }

}