//Author:Karolina Miśkiewicz #include<iostream> using namespace std; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n, m; cin >> n >> m; int tab[n], nm[n]; for(int i = 0; i < n; i++) { int a; cin >> a; tab[i] = a; if(i == 0) nm[i] = a; else nm[i] = min(nm[i-1],a); } if(m == 2) { int mini = tab[n-1]; int p = -1; for(int i = n-1; i >= 0; i--) { if(nm[i] >= mini) //you cannot go higher after { p = i; break; } else mini = max(tab[i], mini); } if(p == -1) cout << "NIE"; else cout << "TAK" << "\n" << p; } else if(m == 3) // most complicated one, because you can have good option from front or back { int p = -1; int mini = tab[0]; for(int i = 1; i < n; i++) { if(tab[i] <= mini) { p = i; break; } } if(p != -1) { if(p == n-1) cout << "TAK" << "\n" << "1 " << p; else cout << "TAK" << "\n" << p << " " << p+1; } if(p == -1) { mini = tab[n-1]; for(int i = n-2; i >= 0; i--) { if(tab[i] >= mini) { p = i; break; } } if(p == -1) cout << "NIE"; else { cout << "TAK" << "\n"; if(p == 1) cout << p << " 2"; else cout << p - 1 << " " << p; } } } else // we have to just find two numbers where a >= b, then rest is not consequential { int p = -1; for(int i = 1; i < n; i++) { if(tab[i-1] >= tab[i]) { p = i; break; } } if(p != -1) { cout << "TAK" << "\n"; if(p+1 <= m) { for(int i = 1; i < m; i++) cout << i << " "; } else { for(int i = 1; i+2 < m; i++) cout << i << " "; cout << p << " " << p+1; } } else cout << "NIE"; } }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 | //Author:Karolina Miśkiewicz #include<iostream> using namespace std; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n, m; cin >> n >> m; int tab[n], nm[n]; for(int i = 0; i < n; i++) { int a; cin >> a; tab[i] = a; if(i == 0) nm[i] = a; else nm[i] = min(nm[i-1],a); } if(m == 2) { int mini = tab[n-1]; int p = -1; for(int i = n-1; i >= 0; i--) { if(nm[i] >= mini) //you cannot go higher after { p = i; break; } else mini = max(tab[i], mini); } if(p == -1) cout << "NIE"; else cout << "TAK" << "\n" << p; } else if(m == 3) // most complicated one, because you can have good option from front or back { int p = -1; int mini = tab[0]; for(int i = 1; i < n; i++) { if(tab[i] <= mini) { p = i; break; } } if(p != -1) { if(p == n-1) cout << "TAK" << "\n" << "1 " << p; else cout << "TAK" << "\n" << p << " " << p+1; } if(p == -1) { mini = tab[n-1]; for(int i = n-2; i >= 0; i--) { if(tab[i] >= mini) { p = i; break; } } if(p == -1) cout << "NIE"; else { cout << "TAK" << "\n"; if(p == 1) cout << p << " 2"; else cout << p - 1 << " " << p; } } } else // we have to just find two numbers where a >= b, then rest is not consequential { int p = -1; for(int i = 1; i < n; i++) { if(tab[i-1] >= tab[i]) { p = i; break; } } if(p != -1) { cout << "TAK" << "\n"; if(p+1 <= m) { for(int i = 1; i < m; i++) cout << i << " "; } else { for(int i = 1; i+2 < m; i++) cout << i << " "; cout << p << " " << p+1; } } else cout << "NIE"; } } |