# PROBLEM LINK:

**Setter:** Anik Sarker, Ezio Auditore

**Tester:** Teja Vardhan Reddy

**Editorialist:** Taranpreet Singh

# DIFFICULTY:

# PREREQUISITES:

Binary Search, Math, Precomputation.

# PROBLEM:

You have two integers X and Y both initialized to zero. Let’s define an operation as

- Choosing any positive integer P such that P*P > Y
- Set X = P
- Set Y = Y+P*P

We need to answer queries, each query specifying a value V, we need to determine the maximum number of operations we can perform such that at the end, X = V.

# EXPLANATION

Let us make some observations first.

For all operations except last, only the value of Y matters. It is obvious since past operations do not affect the current value of X. The only affected value is Y. For the last operation, we shall always choose the value given in the current query as P.

In every operation except the last operation, it is optimal to choose the smallest P such that P*P > Y.

The reason is that suppose R is the minimum value of P such that P*P > Y, then if we choose say R+1, then Y increases by (R+1)*(R+1) instead of R*R. This forces us to choose the higher value of P in future operations, which, combined with the fact that Y do not decrease, reduces the number of operations. Hence, at every operation, we choose the smallest P such that P*P > Y.

Now, the last thing we can notice is that performing all except the last operation is independent of the value of V as long as V*V > Y holds. So, we can just compute the minimum value of Y after q operations for each q. This sequence is always a strictly increasing sequence, and we need the rightmost position such that value at that position do not exceed V*V. Hence, we can perform a binary search to find out the rightmost position. This position gives the maximum number of operations we can perform. We need to add 1 for the final operation too.

Another solution is to just simulate the process, just by noticing the fact that the growth of the value of Y (By choosing the minimum value of P at each operation) ensures no more than approximately 59 iterations for each query, which also fits the time limit easily.

# TIME COMPLEXITY

Time complexity is O(A+Q*log(A)) or O(Q*A) where A = 59 is the maximum possible answer. (Can add also be O(A*log(MX)+Q*log(A)) if we use binary search to find minimum value of P).

# SOLUTIONS:

## Setter 1 Solution

```
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
const ll INF = 2e9;
vector<ll> vec;
int main(){
ll X = 0;
ll Y = 0;
while(X < INF){
X = sqrtl(Y);
while(X*X <= Y) X++;
while((X-1)*(X-1) > Y) X--;
vec.push_back(X);
Y += X * X;
}
int t;
scanf("%d",&t);
assert(1 <= t <= 1000000);
for(int cs=1; cs<=t; cs++){
ll z;
scanf("%lld",&z);
assert(1 <= z <= 1000000000);
int idx = upper_bound(vec.begin(),vec.end(),z) - vec.begin();
printf("%d\n",idx);
}
}
```

## Setter 2 Solution

```
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n, m, x, y, z;
ll t, cs = 1;
int main()
{
cin >> t;
if(t < 1 || t > 1000000) assert(false);
while(t--){
scanf("%lld", &n);
if(n < 1 || n > 1000000000) assert(false);
ll x = 0, y = 0;
int cc = 0;
while(x <= n){
ll tmp = sqrt(y);
tmp++;
y += tmp * tmp;
x = tmp;
cc++;
}
printf("%d\n", cc - 1);
}
return 0;
}
```

## Tester's Solution

```
//teja349
#include <bits/stdc++.h>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <utility>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <iomanip>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
//setbase - cout << setbase (16); cout << 100 << endl; Prints 64
//setfill - cout << setfill ('x') << setw (5); cout << 77 << endl; prints xxx77
//setprecision - cout << setprecision (14) << f << endl; Prints x.xxxx
//cout.precision(x) cout<<fixed<<val; // prints x digits after decimal in val
using namespace std;
using namespace __gnu_pbds;
#define f(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) f(i,0,n)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define vi vector< int >
#define vl vector< ll >
#define ss second
#define ff first
#define ll long long
#define pii pair< int,int >
#define pll pair< ll,ll >
#define sz(a) a.size()
#define inf (1000*1000*1000+5)
#define all(a) a.begin(),a.end()
#define tri pair<int,pii>
#define vii vector<pii>
#define vll vector<pll>
#define viii vector<tri>
#define mod (1000*1000*1000+7)
#define pqueue priority_queue< int >
#define pdqueue priority_queue< int,vi ,greater< int > >
#define flush fflush(stdout)
#define primeDEN 727999983
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// find_by_order() // order_of_key
typedef tree<
int,
null_type,
less<int>,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
#define int ll
main(){
//std::ios::sync_with_stdio(false); cin.tie(NULL);
int t;
//cin>>t;
scanf("%lld",&t);
int y=0;
int x=0;
vi vec;
//vec.pb(0);
while(x<inf){
x=max((int)sqrt(y)-3,(int)0);
while(x*x<=y)
x++;
y+=x*x;
vec.pb(x);
}
int val;
//cerr<<vec.size()<<endl;
while(t--){
//cin>>x;
scanf("%lld",&x);
val = lower_bound(all(vec),x+1)-vec.begin();
printf("%lld\n",val);
//cout<<lower_bound(all(vec),x+1)-vec.begin()<<endl;
}
return 0;
}
```

## Editorialist's Solution

```
import java.util.*;
import java.io.*;
import java.text.*;
class TWOVRIBL{
//SOLUTION BEGIN
//Into the Hardware Mode
void pre() throws Exception{}
void solve(int TC) throws Exception{
long Y = 0, X = 0;
long[] a = new long[100];int c = 0;
while(true){
X = sq(Y);
//Maximum value of X is 1e9
if(X > 1e9)break;
Y += X*X;
a[c++] = X;
}
for(int q = ni(); q>0; q--){
long x = nl();
int lo = 0, hi = c-1;
while(lo+1 < hi){
int mid = lo+(hi-lo)/2;
if(a[mid] <= x)lo = mid;
else hi = mid;
}
if(a[hi] <= x)pn(hi+1);
else pn(lo+1);
}
}
//Finds P such that P*P > x
long sq(long x){
long lo = 1, hi = (long)1e9+1;
while(lo+1< hi){
long mid = lo+(hi-lo)/2;
if(mid*mid > x)hi = mid;
else lo = mid;
}
if(lo*lo > x)return lo;
return hi;
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
long IINF = (long)1e18, mod = (long)1e9+7;
final int INF = (int)1e9, MX = (int)2e5+5;
DecimalFormat df = new DecimalFormat("0.00000000000");
double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-6;
static boolean multipleTC = false, memory = false, fileIO = false;
FastReader in;PrintWriter out;
void run() throws Exception{
if(fileIO){
in = new FastReader("input.txt");
out = new PrintWriter("output.txt");
}else {
in = new FastReader();
out = new PrintWriter(System.out);
}
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
if(memory)new Thread(null, new Runnable() {public void run(){try{new TWOVRIBL().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
else new TWOVRIBL().run();
}
long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str = "";
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}
```

Feel free to share your approach, if you want to. (even if its same ) . Suggestions are welcomed as always had been.