HackerEarth Summation program problem solution

In this HackerEarth Summation program problem solution you are given a number N. you are required to determine the value of the following function:

long long int solve(N)
{
    ans=0;
    for(i=1;i<=N;i++)
    ans+=(N/i);
    return ans;
}

All divisions are integer divisions(i.e. N/i is actually floor(N/i)).


HackerEarth Summation program problem solution


HackerEarth Summation program problem solution.

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll n,ans,t;
void solve()
{
ll i,j,k;
cin>>t;
while(t--)
{
cin>>n;
ans=0;
j=sqrt(n);
for(i=1;i<=j;i++)
{
ans+=(n/i);
}
for(i=1;i<=j;i++)
{
ll lo,hi;
hi=n/i;
lo=n/(i+1);
lo=max(lo,j);
hi=max(hi,j);
ans+=i*(hi-lo);
}
cout<<ans<<"\n";
}
}
int main()
{

solve();
return 0;
}

second solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class FloorSum {
static class FastReader {
BufferedReader br;
StringTokenizer st;

public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}

String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}

int nextInt() {
return Integer.parseInt(next());
}

long nextLong() {
return Long.parseLong(next());
}

double nextDouble() {
return Double.parseDouble(next());
}

String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}

public static void main(String[] args) {

FastReader fr = new FastReader();
int t = fr.nextInt();
assertion(t >= 1 && t <= 100);
for (int i = 0; i < t; i++) {
long n = fr.nextLong();
assertion(n >= 1 && n <= 1e12);
System.out.println(evaluate(n));
}

}

private static long evaluate(long n) {
long ans = 0;
if (n == 1) {
return 1;
}
ans += 1 + n;
for (long i = 2; i * i <= n; i++) {
if (i != n / i) {
ans += (i + n / i);
} else {
ans += i;
}
long l = (n / i) + 1;
long h = (n / (i - 1)) - 1;
if (l <= h) {
ans += (h-l+1)*(i-1);
}

}
return ans;
}

private static long evaluate_brute(long n) {
long ans = 0;
for (int i = 1; i < n; i++) {
ans += (n / i);
}
return ans;
}

private static void assertion(boolean condition) {
if (!condition)
throw new AssertionError();
}
}


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