Question Name: Bishu and Soldiers

#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n,i;
    cin>>n;
    vector<int>v(n);
    for(i=0;i<n;i++)
    cin>>v[i];
    sort(v.begin(),v.end());
    int q;
    cin>>q;
    while(q--)
    {
        int p,sum=0,j=0;
      cin>>p;
      for(i=0;i<n;i++)
      {
          if(v[i]>p)
          break;
          sum+=v[i];
          j++;
        }  
        cout<<j<<" "<<sum<<"\n";
    }
    return 0;
}

• Problem Description
  Bishu went to fight for Coding Club. There were N soldiers with various powers. There will be Q rounds to fight and in each round Bishu’s power will be varied. With power M, Bishu can kill all the soldiers whose power is less than or equal to M(<=M).
  
  After each round, All the soldiers who are dead in previous round will reborn.Such that in each round there will be N soldiers to fight. As Bishu is weak in mathematics, help him to count the number of soldiers that he can kill in each round and total sum of their powers.
  
  1<=N<=10000
  
  1<=power of each soldier<=100
  
  1<=Q<=10000
  
  1<=power of bishu<=100
• Test Case 1
  Input (stdin)4
  1 2 3 4
  3
  3
  10
  2
  Expected Output3 6
  4 10
  2 3
• Test Case 2
  Input (stdin)7
  1 2 3 4 5 6 7
  3
  3
  10
  2
  Expected Output3 6
  7 28
  2 3