Codeforces round 1037(A. Packets)

When you have 1 then just for 1=1.so needed coins=1.
in the case of 2 (1 coin& 1 coin).so needed coins=2.
   1=1
   2=1+1
in the case of 3 (1 coin & 2 coin).so needed coins=2.
   1=1
   2=2
   3=1+2
in the case of 4 (1 coin, 1 coin & 2 coin).so needed coins=3.
   1=1
   2=2
   3=1+2
   4=1+1+2
in the case of 5 (1 coin ,2 coin & 2 coin).so needed coins=3.
   1=1
   2=2
   3=1+2
   4=2+2
   5=1+2+2
in the case of 6 (1 coin ,2 coin & 3 coin).so needed coins=3.
   1=1
   2=2
   3=3
   4=1+3
   5=2+3
   6=1+2+3
in the case of 7 (1 coin ,2 coin & 4 coin).so needed coins=3.
   1=1
   2=2
   3=1+2
   4=4
   5=1+4
   6=2+4
   7=1+2+4
in the case of 8 (1 coin ,2 coin,2 coin & 3 coin).so needed coins=4.
   1=1
   2=2
   3=1+2
   4=2+2
   5=1+2+2
   6=1+2+3
   7=2+2+3
   8=1+2+2+3

///Bismillahir Rahmanir Rahim
///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
  angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;

int main()
{
    ll n,cnt=0;
    cin>>n;
    while(n!=0) n/=2,cnt++;
    cout<<cnt<<endl;
    return 0;
}
///Alhamdulillah

Problem Link:https://codeforces.com/problemset/problem/1037/A

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