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LIS_and_LDS.cpp
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LIS_and_LDS.cpp
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//
// LIS and LDS
//
// verified:
// ARC 091 E - LISDL
// https://beta.atcoder.jp/contests/arc091/tasks/arc091_c
//
/*
0 〜 N-1 の順列であって
・LIS (最長増加部分列) の長さが A
・LDS (最長減少部分列) の長さが B
となるものを 1 つ求める
そのようなものが存在する必要十分条件は
・A + B <= N+1
・AB >= N
*/
#include <iostream>
#include <vector>
using namespace std;
vector<long long> LISLDS(long long N, long long A, long long B) {
vector<long long> res;
for (long long i = B-1; i >= 0; --i) res.push_back(i);
if (A == 1) return res;
long long rem = N - B;
long long p = rem / (A-1);
long long r = rem % (A-1);
long long b = r;
long long c = A-1-r;
for (int i = 0; i < b; ++i) {
int size = (int)res.size();
for (int j = 0; j < p+1; ++j) {
res.push_back(size + p - j);
}
}
for (int i = 0; i < c; ++i) {
int size = (int)res.size();
for (int j = 0; j < p; ++j) {
res.push_back(size + p-1 - j);
}
}
return res;
}
//------------------------------//
// Examples
//------------------------------//
int main() {
long long N, A, B; cin >> N >> A >> B;
if (A + B > N+1 || A * B < N) cout << -1 << endl;
else {
vector<long long> res = LISLDS(N, A, B);
for (int i = 0; i < res.size(); ++i) {
cout << res[i] + 1;
if (i != (int)res.size() - 1) cout << " ";
}
cout << endl;
}
}