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

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