Decimal to Irrational fraction approximation - algorithm

I have implemented an algorithm for floating point decimal to rational fraction approximation (example: 0.333 -> 1/3) and now I wonder, is there a way to find an irrational number which satisfies the condition. For example, given the input 0.282842712474 I want the result to be sqrt(2)/5 and not 431827/1526739 which my algorithm produces. The only condition is that the first digits of the result (converted back to floating point) should be the digits of the input, the rest doesn't matter. Thanks in advance!

I came up with solution, that from given set of possible denominators and nominators finds best approximation of given number.
For example this set can contain all numbers that can be created by:
1 <= radicand <= 100000
1 <= root_index <= 20
If set has N elements, than this solution finds best approximation in O(N log N).
In this solution X represents denominator and Y nominator.
sort numbers from set
for each number X from set:
using binary find smallest Y such that Y/X >= input_number
compare Y/X with currently best approximation of input_number
I couldn't resist and I implemented it:
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
struct Number {
// number value
double value;
// number representation
int root_index;
int radicand;
Number(){}
Number(double value, int root_index, int radicand)
: value(value), root_index(root_index), radicand(radicand) {}
bool operator < (const Number& rhs) const {
// in case of equal numbers, i want smaller radicand first
if (fabs(value - rhs.value) < 1e-12) return radicand < rhs.radicand;
return value < rhs.value;
}
void print() const {
if (value - (int)value < 1e-12) printf("%.0f", value);
else printf("sqrt_%d(%d)",root_index, radicand);
}
};
std::vector<Number> numbers;
double best_result = 1e100;
Number best_numerator;
Number best_denominator;
double input;
void compare_approximpation(const Number& numerator, const Number& denominator) {
double value = numerator.value / denominator.value;
if (fabs(value - input) < fabs(best_result - input)) {
best_result = value;
best_numerator = numerator;
best_denominator = denominator;
}
}
int main() {
const int NUMBER_LIMIT = 100000;
const int ROOT_LIMIT = 20;
// only numbers created by this loops will be used
// as numerator and denominator
for(int i=1; i<=ROOT_LIMIT; i++) {
for(int j=1; j<=NUMBER_LIMIT; j++) {
double value = pow(j, 1.0 /i);
numbers.push_back(Number(value, i, j));
}
}
sort(numbers.begin(), numbers.end());
scanf("%lf",&input);
int numerator_index = 0;
for(int denominator_index=0; denominator_index<numbers.size(); denominator_index++) {
// you were interested only in integral denominators
if (numbers[denominator_index].root_index == 1) {
// i use simple sweeping technique instead of binary search (its faster)
while(numerator_index < numbers.size() && numbers[numerator_index].root_index &&
numbers[numerator_index].value / numbers[denominator_index].value <= input) {
numerator_index++;
}
// comparing approximations
compare_approximpation(numbers[numerator_index], numbers[denominator_index]);
if (numerator_index > 0) {
compare_approximpation(numbers[numerator_index - 1], numbers[denominator_index]);
}
}
}
printf("Best approximation %.12lf = ", best_numerator.value / best_denominator.value);
best_numerator.print();
printf(" / ");
best_denominator.print();
printf("\n");
}

Related

Extracting patterns from time-series

I have a time-series, which essentially amounts to some instrument recording the current time whenever it makes a "detection". The sampling rate is therefore not in constant time, however we can treat it as such by "re-sampling", relying on the fact that the detections are made reliably and we can simply insert 0's to "fill in" the gaps. This will be important later...
The instrument should detect the "signals" sent by another, nearby instrument. This second instrument emits a signal at some unknown period, T (e.g. 1 signal per second), with a "jitter" likely on the order of a few tenths of a percent of the period.
My goal is to determine this period (or frequency, if you like) using only the timestamps recorded by the "detecting" instrument. Unfortunately, however, the detector is flooded with noise, and a significant amount (I estimate 97-98%) of "detections" (and therefore "points" in the time-series) are due to noise. Therefore, extracting the period will require more careful analysis.
My first thought was to simply feed the time series into an FFT algorithm (I'm using FFTW/DHT), however this wasn't particularly enlightening. I've also tried my own (admittedly rather crude) algorithm, which simply computed a cross-correlation of the series with "clean" series of increasing period. I didn't get very far with this, either, and there are quite a handful of details to consider (phase, etc).
It occurs to me that something like this must've been done before, and surely there's a "nice" way to accomplish it.
Here's my approach. Given a period, we can score it using a dynamic program to find the subsequence of detection times that includes the first and last detection and maximizes the sum of gap log-likelihoods, where the gap log-likelihood is defined as minus the square of the difference of the gap and the period (Gaussian jitter model).
If we have approximately the right period, then we can get a very good gap sequence (some weirdness at the beginning and end and wherever there is a missed detection, but this is OK).
If we have the wrong period, then we end up with basically exponential jitter, which has low log-likelihood.
The C++ below generates fake detection times with a planted period and then searches over periods. Scores are normalized by a (bad) estimate of the score for Poisson noise, so wrong periods score about 0.4. See the plot below.
#include <algorithm>
#include <cmath>
#include <iostream>
#include <limits>
#include <random>
#include <vector>
namespace {
static constexpr double kFalseNegativeRate = 0.01;
static constexpr double kCoefficientOfVariation = 0.003;
static constexpr int kSignals = 6000;
static constexpr int kNoiseToSignalRatio = 50;
template <class URNG>
std::vector<double> FakeTimes(URNG &g, const double period) {
std::vector<double> times;
std::bernoulli_distribution false_negative(kFalseNegativeRate);
std::uniform_real_distribution<double> phase(0, period);
double signal = phase(g);
std::normal_distribution<double> interval(period,
kCoefficientOfVariation * period);
std::uniform_real_distribution<double> noise(0, kSignals * period);
for (int i = 0; i < kSignals; i++) {
if (!false_negative(g)) {
times.push_back(signal);
}
signal += interval(g);
for (double j = 0; j < kNoiseToSignalRatio; j++) {
times.push_back(noise(g));
}
}
std::sort(times.begin(), times.end());
return times;
}
constexpr double Square(const double x) { return x * x; }
struct Subsequence {
double score;
int previous;
};
struct Result {
double score = std::numeric_limits<double>::quiet_NaN();
double median_interval = std::numeric_limits<double>::quiet_NaN();
};
Result Score(const std::vector<double> &times, const double period) {
if (times.empty() || !std::is_sorted(times.begin(), times.end())) {
return {};
}
std::vector<Subsequence> bests;
bests.reserve(times.size());
bests.push_back({0, -1});
for (int i = 1; i < times.size(); i++) {
Subsequence best = {std::numeric_limits<double>::infinity(), -1};
for (int j = i - 1; j > -1; j--) {
const double difference = times[i] - times[j];
const double penalty = Square(difference - period);
if (difference >= period && penalty >= best.score) {
break;
}
const Subsequence candidate = {bests[j].score + penalty, j};
if (candidate.score < best.score) {
best = candidate;
}
}
bests.push_back(best);
}
std::vector<double> intervals;
int i = bests.size() - 1;
while (true) {
int previous_i = bests[i].previous;
if (previous_i < 0) {
break;
}
intervals.push_back(times[i] - times[previous_i]);
i = previous_i;
}
if (intervals.empty()) {
return {};
}
const double duration = times.back() - times.front();
// The rate is doubled because we can look for a time in either direction.
const double rate = 2 * (times.size() - 1) / duration;
// Mean of the square of an exponential distribution with the given rate.
const double mean_square = 2 / Square(rate);
const double score = bests.back().score / (intervals.size() * mean_square);
const auto median_interval = intervals.begin() + intervals.size() / 2;
std::nth_element(intervals.begin(), median_interval, intervals.end());
return {score, *median_interval};
}
} // namespace
int main() {
std::default_random_engine g;
const auto times = FakeTimes(g, std::sqrt(2));
for (int i = 0; i < 2000; i++) {
const double period = std::pow(1.001, i) / 3;
const Result result = Score(times, period);
std::cout << period << ' ' << result.score << ' ' << result.median_interval
<< std::endl;
}
}

why the constraints reduce the accepted test cases

this is a code i wrote to solve a problem on HackerRank "Recursive Digit Sum"
https://www.hackerrank.com/challenges/recursive-digit-sum
,the code suppose to take two digits as inputs (n,k) to calculate the super digit p.
p is created when k is concatenated n times thats if, k=148 and n=3
p=148148148
sumdigit(P) = sumdigit(148148148)
= sumdigit(1+4+8+1+4+8+1+4+8)
= sumdigit(39)
= sumdigit(3+9)
= sumdigit(12)
= sumdigit(1+2)
= sumdigit(3)
= 3.
the Constraints
1<=n<10^(100000)
1<=k<=10^5
#include <stdio.h>
#include <math.h>
unsigned long int SumDigits(unsigned long int i) {
if (i < 10) {
return i;
}
else {
return i%10 + SumDigits(i/10);
}
}
int main() {
unsigned long int n,k,pos=0;
scanf("%ld %ld",&k,&n);
/**i would put the constraint however it reduces the accepted test cases*/
// if(k>=1&&k<pow(10,100000)&&n>=1&&n<pow(10,5)){
for(unsigned long int i=0;i<n;i++){
pos+=k;
k=k*( unsigned long int)pow(10,n);
}
while(pos>=10){
pos=SumDigits(pos);
}
printf("%ld\n",pos);
//}
return 0;
}

Find all anagrams in a string O(n) solution

Here is the problem:
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Input: s: "cbaebabacd" p: "abc"
Output: [0, 6]
Input: s: "abab" p: "ab"
Output: [0, 1, 2]
Here is my solution
vector<int> findAnagrams(string s, string p) {
vector<int> res, s_map(26,0), p_map(26,0);
int s_len = s.size();
int p_len = p.size();
if (s_len < p_len) return res;
for (int i = 0; i < p_len; i++) {
++s_map[s[i] - 'a'];
++p_map[p[i] - 'a'];
}
if (s_map == p_map)
res.push_back(0);
for (int i = p_len; i < s_len; i++) {
++s_map[s[i] - 'a'];
--s_map[s[i - p_len] - 'a'];
if (s_map == p_map)
res.push_back(i - p_len + 1);
}
return res;
}
However, I think it is O(n^2) solution because I have to compare vectors s_map and p_map.
Does a O(n) solution exist for this problem?
lets say p has size n.
lets say you have an array A of size 26 that is filled with the number of a,b,c,... which p contains.
then you create a new array B of size 26 filled with 0.
lets call the given (big) string s.
first of all you initialize B with the number of a,b,c,... in the first n chars of s.
then you iterate through each word of size n in s always updating B to fit this n-sized word.
always B matches A you will have an index where we have an anagram.
to change B from one n-sized word to another, notice you just have to remove in B the first char of the previous word and add the new char of the next word.
Look at the example:
Input
s: "cbaebabacd"
p: "abc" n = 3 (size of p)
A = {1, 1, 1, 0, 0, 0, ... } // p contains just 1a, 1b and 1c.
B = {1, 1, 1, 0, 0, 0, ... } // initially, the first n-sized word contains this.
compare(A,B)
for i = n; i < size of s; i++ {
B[ s[i-n] ]--;
B[ s[ i ] ]++;
compare(A,B)
}
and suppose that compare(A,B) prints the index always A matches B.
the total complexity will be:
first fill of A = O(size of p)
first fill of B = O(size of s)
first comparison = O(26)
for-loop = |s| * (2 + O(26)) = |s| * O(28) = O(28|s|) = O(size of s)
____________________________________________________________________
2 * O(size of s) + O(size of p) + O(26)
which is linear in size of s.
Your solution is the O(n) solution. The size of the s_map and p_map vectors is a constant (26) that doesn't depend on n. So the comparison between s_map and p_map takes a constant amount of time regardless of how big n is.
Your solution takes about 26 * n integer comparisons to complete, which is O(n).
// In papers on string searching algorithms, the alphabet is often
// called Sigma, and it is often not considered a constant. Your
// algorthm works in (Sigma * n) time, where n is the length of the
// longer string. Below is an algorithm that works in O(n) time even
// when Sigma is too large to make an array of size Sigma, as long as
// values from Sigma are a constant number of "machine words".
// This solution works in O(n) time "with high probability", meaning
// that for all c > 2 the probability that the algorithm takes more
// than c*n time is 1-o(n^-c). This is a looser bound than O(n)
// worst-cast because it uses hash tables, which depend on randomness.
#include <functional>
#include <iostream>
#include <type_traits>
#include <vector>
#include <unordered_map>
#include <vector>
using namespace std;
// Finding a needle in a haystack. This works for any iterable type
// whose members can be stored as keys of an unordered_map.
template <typename T>
vector<size_t> AnagramLocations(const T& needle, const T& haystack) {
// Think of a contiguous region of an ordered container as
// representing a function f with the domain being the type of item
// stored in the container and the codomain being the natural
// numbers. We say that f(x) = n when there are n x's in the
// contiguous region.
//
// Then two contiguous regions are anagrams when they have the same
// function. We can track how close they are to being anagrams by
// subtracting one function from the other, pointwise. When that
// difference is uniformly 0, then the regions are anagrams.
unordered_map<remove_const_t<remove_reference_t<decltype(*needle.begin())>>,
intmax_t> difference;
// As we iterate through the haystack, we track the lead (part
// closest to the end) and lag (part closest to the beginning) of a
// contiguous region in the haystack. When we move the region
// forward by one, one part of the function f is increased by +1 and
// one part is decreased by -1, so the same is true of difference.
auto lag = haystack.begin(), lead = haystack.begin();
// To compare difference to the uniformly-zero function in O(1)
// time, we make sure it does not contain any points that map to
// 0. The the property of being uniformly zero is the same as the
// property of having an empty difference.
const auto find = [&](const auto& x) {
difference[x]++;
if (0 == difference[x]) difference.erase(x);
};
const auto lose = [&](const auto& x) {
difference[x]--;
if (0 == difference[x]) difference.erase(x);
};
vector<size_t> result;
// First we initialize the difference with the first needle.size()
// items from both needle and haystack.
for (const auto& x : needle) {
lose(x);
find(*lead);
++lead;
if (lead == haystack.end()) return result;
}
size_t i = 0;
if (difference.empty()) result.push_back(i++);
// Now we iterate through the haystack with lead, lag, and i (the
// position of lag) updating difference in O(1) time at each spot.
for (; lead != haystack.end(); ++lead, ++lag, ++i) {
find(*lead);
lose(*lag);
if (difference.empty()) result.push_back(i);
}
return result;
}
int main() {
string needle, haystack;
cin >> needle >> haystack;
const auto result = AnagramLocations(needle, haystack);
for (auto x : result) cout << x << ' ';
}
import java.util.*;
public class FindAllAnagramsInAString_438{
public static void main(String[] args){
String s="abab";
String p="ab";
// String s="cbaebabacd";
// String p="abc";
System.out.println(findAnagrams(s,p));
}
public static List<Integer> findAnagrams(String s, String p) {
int i=0;
int j=p.length();
List<Integer> list=new ArrayList<>();
while(j<=s.length()){
//System.out.println("Substring >>"+s.substring(i,j));
if(isAnamgram(s.substring(i,j),p)){
list.add(i);
}
i++;
j++;
}
return list;
}
public static boolean isAnamgram(String s,String p){
HashMap<Character,Integer> map=new HashMap<>();
if(s.length()!=p.length()) return false;
for(int i=0;i<s.length();i++){
char chs=s.charAt(i);
char chp=p.charAt(i);
map.put(chs,map.getOrDefault(chs,0)+1);
map.put(chp,map.getOrDefault(chp,0)-1);
}
for(int val:map.values()){
if(val!=0) return false;
}
return true;
}
}

Parallel radix sort with virtual memory and write-combining

I'm attempting to implement the variant of parallel radix sort described in http://arxiv.org/pdf/1008.2849v2.pdf (Algorithm 2), but my C++ implementation (for 4 digits in base 10) contains a bug that I'm unable to locate.
For debugging purposes I'm using no parallelism, but the code should still sort correctly.
For instance the line arr.at(i) = item accesses indices outside its bounds in the following
std::vector<int> v = {4612, 4598};
radix_sort2(v);
My implementation is as follows
#include <set>
#include <array>
#include <vector>
void radix_sort2(std::vector<int>& arr) {
std::array<std::set<int>, 10> buckets3;
for (const int item : arr) {
int d = item / 1000;
buckets3.at(d).insert(item);
}
//Prefix sum
std::array<int, 10> outputIndices;
outputIndices.at(0) = 0;
for (int i = 1; i < 10; ++i) {
outputIndices.at(i) = outputIndices.at(i - 1) +
buckets3.at(i - 1).size();
}
for (const auto& bucket3 : buckets3) {
std::array<std::set<int>, 10> buckets0, buckets1;
std::array<int, 10> histogram2 = {};
for (const int item : bucket3) {
int d = item % 10;
buckets0.at(d).insert(item);
}
for (const auto& bucket0 : buckets0) {
for (const int item : bucket0) {
int d = (item / 10) % 10;
buckets1.at(d).insert(item);
int d2 = (item / 100) % 10;
++histogram2.at(d2);
}
}
for (const auto& bucket1 : buckets1) {
for (const int item : bucket1) {
int d = (item / 100) % 10;
int i = outputIndices.at(d) + histogram2.at(d);
++histogram2.at(d);
arr.at(i) = item;
}
}
}
}
Can anyone spot my mistake?
I took at look at the paper you linked. You haven't made any mistakes, none that I can see. In fact, in my estimation, you corrected a mistake in the algorithm.
I wrote out the algorithm and ended up with the exact same problem as you did. After reviewing Algorithm 2, either I woefully mis-understand how it is supposed to work, or it is flawed. There are at least a couple of problems with the algorithm, specifically revolving around outputIndices, and histogram2.
Looking at the algorithm, the final index of an item is determined by the counting sort stored in outputIndices. (lets ignore the histogram for now).
If you had an inital array of numbers {0100, 0103, 0102, 0101} The prefix sum of that would be 4.
The algorithm makes no indication I can determine to lag the result by 1. That being said, in order for the algorithm to work the way they intend, it does have to be lagged, so, moving on.
Now, the prefix sums are 0, 4, 4.... The algorithm doesn't use the MSD as the index into the outputIndices array, it uses "MSD - 1"; So taking 1 as the index into the array, the starting index for the first item without the histogram is 4! Outside the array on the first try.
The outputIndices is built with the MSD, it makes sense for it to be accessed by MSD.
Further, even if you tweak the algorithm to correctly to use the MSD into the outputIndices, it still won't sort correctly. With your initial inputs (swapped) {4598, 4612}, they will stay in that order. They are sorted (locally) as if they are 2 digit numbers. If you increase it to have other numbers not starting with 4, they will be globally, sorted, but the local sort is never finished.
According to the paper the goal is to use the histogram to do that, but I don't see that happening.
Ultimately, I'm assuming, what you want is an algorithm that works the way described. I've modified the algorithm, keeping with the overall stated goal of the paper of using the MSD to do a global sort, and the rest of the digits by reverse LSD.
I don't think these changes should have any impact on your desire to parallel-ize the function.
void radix_sort2(std::vector<int>& arr)
{
std::array<std::vector<int>, 10> buckets3;
for (const int item : arr)
{
int d = item / 1000;
buckets3.at(d).push_back(item);
}
//Prefix sum
std::array<int, 10> outputIndices;
outputIndices.at(0) = 0;
for (int i = 1; i < 10; ++i)
{
outputIndices.at(i) = outputIndices.at(i - 1) + buckets3.at(i - 1).size();
}
for (const auto& bucket3 : buckets3)
{
if (bucket3.size() <= 0)
continue;
std::array<std::vector<int>, 10> buckets0, buckets1, buckets2;
for (const int item : bucket3)
buckets0.at(item % 10).push_back(item);
for (const auto& bucket0 : buckets0)
for (const int item : bucket0)
buckets1.at((item / 10) % 10).push_back(item);
for (const auto& bucket1 : buckets1)
for (const int item : bucket1)
buckets2.at((item / 100) % 10).push_back(item);
int count = 0;
for (const auto& bucket2 : buckets2)
{
for (const int item : bucket2)
{
int d = (item / 1000) % 10;
int i = outputIndices.at(d) + count;
++count;
arr.at(i) = item;
}
}
}
}
For extensiblility, it would probably make sense to create a helper function that does the local sorting. You should be able to extend it to handle any number of digit numbers that way.

Finding an efficient algorithm

You are developing a smartphone app. You have a list of potential
customers for your app. Each customer has a budget and will buy the app at
your declared price if and only if the price is less than or equal to the
customer's budget.
You want to fix a price so that the revenue you earn from the app is
maximized. Find this maximum possible revenue.
For instance, suppose you have 4 potential customers and their budgets are
30, 20, 53 and 14. In this case, the maximum revenue you can get is 60.
**Input format**
Line 1 : N, the total number of potential customers.
Lines 2 to N+1: Each line has the budget of a potential customer.
**Output format**
The output consists of a single integer, the maximum possible revenue you
can earn from selling your app.
Also, upper bound on N is 5*(10^5) and upper bound on each customer's budget is 10^8.
This is a problem I'm trying to solve . My strategy was to sort the list of budgets and then multiply each of those with its position-index in the sequence - and then print the max of the resulting sequence. However this seems to be quite time-inefficient (at least in the way I'm implementing it - I've attached the code for reference). My upper bound on time is 2 seconds. Can anyone help me find a
more time-efficient algorithm (or possibly a more efficient way to implement my algorithm) ?
Here is my solution :
#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
using namespace std;
long long max(long long[],long long);
void quickSortIterative(long long[],long long,long long);
long long partition(long long[],long long,long long);
void swap(long long*,long long*);
int main(){
long long n,k=1;
scanf("%lld",&n);
if(n<1 || n > 5*((long long)pow(10,5))){
exit(0);
}
long long budget[n],aux[n];
for(long long i=0;i<n;i++){
scanf("%lld",&budget[i]);
if(budget[i]<1 || budget[i] > (long long)pow(10,8)){
exit(0);
}
}
quickSortIterative(budget,0,n-1);
for(long long j=n-1;j>=0;j--){
aux[j] = budget[j]*k;
k++;
}
cout<<max(aux,n);
return 0;
}
long long partition (long long arr[], long long l, long long h){
long long x = arr[h];
long long i = (l - 1);
for (long long j = l; j <= h- 1; j++)
{
if (arr[j] <= x)
{
i++;
swap (&arr[i], &arr[j]);
}
}
swap (&arr[i + 1], &arr[h]);
return (i + 1);
}
void swap ( long long* a, long long* b ){
long long t = *a;
*a = *b;
*b = t;
}
void quickSortIterative(long long arr[], long long l, long long h){
long long stack[ h - l + 1 ];
long long top = -1;
stack[ ++top ] = l;
stack[ ++top ] = h;
while ( top >= 0 ){
h = stack[ top-- ];
l = stack[ top-- ];
long long p = partition( arr, l, h );
if ( p-1 > l ){
stack[ ++top ] = l;
stack[ ++top ] = p - 1;
}
if ( p+1 < h ){
stack[ ++top ] = p + 1;
stack[ ++top ] = h;
}
}
}
long long max(long long arr[],long long length){
long long max = arr[0];
for(long long i=1;i<length;i++){
if(arr[i]>max){
max=arr[i];
}
}
return max;
}
Quicksort can take O(n^2) time for certain sequences (often already sorted sequences are bad).
I would recommend you try using a sorting approach with guaranteed O(nlogn) performance (e.g. heapsort or mergesort). Alternatively, you may well find that using the sort routines in the standard library will give better performance than your version.
You might use qsort in C or std::sort in C++, which is most likely faster than your own code.
Also, your "stack" array will cause you trouble if the difference h - l is large.
I have used STL library function sort() of C++. It's time complexity is O(nlogn). Here, you just need to sort the given array and check from maximum value to minimum value for given solution. It is O(n) after sorting.
My code which cleared all the test cases :
#include <algorithm>
#include <stdio.h>
#include <cmath>
#include <iostream>
using namespace std;
int main(){
long long n, a[1000000], max;
int i, j;
cin>>n;
for(i = 0; i < n; i++){
cin>>a[i];
}
sort(a, a + n);
max = a[n - 1];
for(i = n - 2; i >= 0; i--){
//printf("%lld ", a[i]);
if(max < (a[i] * (n - i)))
max = a[i] * (n - i);
}
cout<<max<<endl;
return 0;
}
I dont know if my answer is right or wrong please point out mistakes if there is any
#include<stdio.h>
void main()
{
register int i,j;
long long int n,revenue;
scanf("%Ld",&n);
long long int a[n];
for(i=0;i<n;i++)
scanf("%Ld",&a[i]);
for (i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(a[i]>a[j])
{
a[i]=a[i]+a[j];
a[j]=a[i]-a[j];
a[i]=a[i]-a[j];
}
}
}
for(i=0;i<n;i++)
a[i]=(n-i)*a[i];
revenue=0;
for(i=0;i<n;i++)
{
if(revenue<a[i])
revenue=a[i];
}
printf("%Ld\n",revenue);
}
passed all the test cases
n=int(input())
r=[]
for _ in range(n):
m=int(input())
r.append(m)
m=[]
r.sort()
l=len(r)
for i in range(l):
m.append((l-i)*r[i])
print(max(m))
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main() {
// your code goes here
long long n;
std::cin >> n;
long long a[n];
for(long long i=0;i<n;i++)
{
std::cin >> a[i];
}
sort(a,a+n);
long long max=LONG_MIN,count;
for(long long i=0;i<n;i++)
{
if(a[i]*(n-i)>max)
{
max=a[i]*(n-i);
}
}
std::cout << max << std::endl;
return 0;
}
The following solution is in C programming Language.
The Approach is:
Input the number of customers.
Input the budgets of customers.
Sort the budget.
Assign revenue=0
Iterate through the budget and Multiply the particular budget with the remaining budget values.
If the previous-revenue < new-revenue. assign the new-revenue to revenue variable.
The code is as follows:
#include <stdio.h>
int main(void) {
int i,j,noOfCustomer;
scanf("%d",&noOfCustomer);
long long int budgetOfCustomer[noOfCustomer],maximumRevenue=0;
for(i=0;i<noOfCustomer;i++)
{
scanf("%Ld",&budgetOfCustomer[i]);
}
for(i=0;i<noOfCustomer;i++)
{
for(j=i+1;j<noOfCustomer;j++)
{
if(budgetOfCustomer[i]>budgetOfCustomer[j])
{
budgetOfCustomer[i]=budgetOfCustomer[i] + budgetOfCustomer[j];
budgetOfCustomer[j]=budgetOfCustomer[i] - budgetOfCustomer[j];
budgetOfCustomer[i]=budgetOfCustomer[i] - budgetOfCustomer[j];
}
}
}
for(i=0;i<noOfCustomer;i++)
{
budgetOfCustomer[i]=budgetOfCustomer[i]*(noOfCustomer-i);
}
for(i=0;i<noOfCustomer;i++)
{
if(maximumRevenue<budgetOfCustomer[i])
maximumRevenue=budgetOfCustomer[i];
}
printf("%Ld\n",maximumRevenue);
return 0;
}

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