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| -rwxr-xr-x | 14_construction_problems/a.out | bin | 0 -> 16048 bytes | |||
| -rw-r--r-- | 14_construction_problems/inverse_inversions_2214.cpp | 22 |
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diff --git a/14_construction_problems/a.out b/14_construction_problems/a.out new file mode 100755 index 0000000..8f05df6 --- /dev/null +++ b/14_construction_problems/a.out | |||
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diff --git a/14_construction_problems/inverse_inversions_2214.cpp b/14_construction_problems/inverse_inversions_2214.cpp new file mode 100644 index 0000000..7e7ae07 --- /dev/null +++ b/14_construction_problems/inverse_inversions_2214.cpp | |||
| @@ -0,0 +1,22 @@ | |||
| 1 | #include <iostream> | ||
| 2 | #include <utility> | ||
| 3 | #include <vector> | ||
| 4 | |||
| 5 | // Observation: putting 1 at the n-th position (1-based) generates n-1 | ||
| 6 | // inversions; then, putting 2 at n-2 generates another n-2 inversions; | ||
| 7 | // and so on. | ||
| 8 | // The first step in our algorithm counts how many times we can do this, | ||
| 9 | // and leaves k pointing to the position where we should put the next | ||
| 10 | // element. All other elements are in increasing order. | ||
| 11 | |||
| 12 | int main() { | ||
| 13 | long long n, k, p, l, m, j, i{0}; | ||
| 14 | std::cin >> n >> k; | ||
| 15 | std::vector<long long> a(n, 0); | ||
| 16 | |||
| 17 | for (j = 1, p = 0; p < n-1 && k >= n-j; p++, j++) k -= n-j; | ||
| 18 | |||
| 19 | for (m = p+1, l = p+2, i = 0; i < n; i++) | ||
| 20 | std::cout << ((n-i <= p || i == k) ? m-- : l++) << " "; | ||
| 21 | std::cout << "\n"; | ||
| 22 | } | ||
