# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
985119 | ShauryaTheShep | Permutation (APIO22_perm) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric> // Include this header for iota
using namespace std;
// Function to count the number of increasing subsequences
int countIncreasingSubsequences(const vector<int>& permutation) {
int n = permutation.size();
vector<int> dp(n, 1); // dp[i] will store the number of increasing subsequences ending at index i
int totalCount = 0;
// Calculate the number of increasing subsequences ending at each index
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (permutation[j] < permutation[i]) {
dp[i] += dp[j];
}
}
totalCount += dp[i];
}
return totalCount;
}
// Function to construct a permutation with exactly k increasing subsequences
vector<int> construct_permutation(int64_t k) {
int n = 2;
vector<int> permutation;
while (true) {
permutation.resize(n);
iota(permutation.begin(), permutation.end(), 1); // Create permutation starting from 1
do {
if (countIncreasingSubsequences(permutation) == k) {
return permutation;
}
} while (next_permutation(permutation.begin(), permutation.end()));
++n;
}
return {}; // Should not reach here if k is within valid range
}