en.md

LOJ 1053 - False Ordering

In this problem, you will be given T test cases. Each test case contains an integer N, the value of which is 1000 at max. Your task is to print out the N^th integer after arranging all the integers from 1 to 1000 following certain conditions.

Conditions: In the arrangement, the integer x will be placed before the integer y if:

• d(x) < d(y)
• x > y when d(x) = d(y)

Note: d(n) denotes the number of divisors of the integer n

One approach to solve the problem is to pre-compute the arrangement and afterward, to print out the answer in O(1) time complexity. This can be achieved by:

1. Calculating the number of divisors of the integers using modified sieve technique
2. Computing the order of integers with the help of custom comparison function of C++

This is a very good problem to help you learn custom comparison in C++. However, some people might find it difficult to get an Accepted verdict. The reason could be one of the following:

1. Errors in calculating d(n)
2. Incorrect comparison conditions
3. Failing to provide proper output format
4. Missing new line on each line

If you are still stuck with this problem, check the code below:

C++

---

#include <bits/stdc++.h>
using namespace std;

#define f first
#define s second
typedef pair<int, int> pii;
const int MX = 1e3;

pii d[MX];

bool order_cond (pii x, pii y) {
  if (x.s != y.s)
    return x.s < y.s;
  return x.f > y.f;
}

void pre_compute() {
  for (int i = 1; i <= MX; i++) {
    d[i - 1].f = i;
    for (int j = i; j <= MX; j += i) {
      d[j - 1].s++;
    }
  }
  sort(d, d + MX, order_cond);
}

int main() {
  int T, N;
  pre_compute();
  cin >> T;
  for(int i = 1; i <= T; i++) {
    cin >> N;
    cout << "Case " << i << ": " << d[N - 1].f << endl;
  }
  return 0;
}