#include <cstdio> #include <vector> #include <unordered_map> #include <set> using namespace std; #define INF 1000000000000000009LL struct pair_hash { template <class T1, class T2> std::size_t operator () (const std::pair<T1,T2> &p) const { auto h1 = std::hash<T1>{}(p.first); auto h2 = std::hash<T2>{}(p.second); return (h1 * 937ull) ^ h2; } }; unordered_map<pair<int, long long>, long long, pair_hash> c; // In how many ways we can get l inversions in n-permutation. long long get(long long n, long long l){ long long mx = n*(n-1)/2; long long half = mx/2; if (l>half) l = mx - l; if (l>200) return INF; if (l>20 && n>100) return INF; if (l>8 && n>1000) return INF; if (l==0) return 1; if (n==1) return 0; pair<int, long long> pp(n, l); auto it = c.find(pp); if (it != c.end()) return it->second; long long sum = 0; long long mm = n-1; if (l<mm) mm = l; for (long long i = 0; i <= mm; i++) { sum += get(n-1, l-i); if (sum > INF) { sum = INF; break; } } return c[pp] = sum; } set<int> remaining; void init(int n){ for(int i=0; i<n; i++){ remaining.insert(i+1); } } int nth(size_t n){ if(n>remaining.size()/2){ //if(0){ n = remaining.size() - n - 1; for(auto it = remaining.rbegin(); it!=remaining.rend(); it++){ if(n-- == 0){ int r = *it; remaining.erase(--it.base()); return r; } } } else{ for(auto it = remaining.begin(); it!=remaining.end(); it++){ if(n-- == 0){ int r = *it; remaining.erase(it); return r; } } } return -1; } void test(){ for(long long n=1; n<200000; n++){ int x=-1; for(int l=0; l<=n*(n-1)/2; l++){ x = l; if(get(n, l)==INF){ break; } } printf("%lld %d\n", n, x); } } int main(){ //test(); return 0; long long n; long long k; scanf("%lld %lld", &n, &k); k--; if((n*(n-1)) % 4){ printf("NIE\n"); return 0; } long long l = n*(n-1)/4; init(n); vector<int> ans; for(int i=0; i<n; i++){ //printf("i=%d, l=%lld, k=%lld\n", i, l, k); bool found=false; long long n1 = n-i-1; long long min_inversions = l - (n1*(n1-1))/2; if(min_inversions<0) min_inversions = 0; int max_inversions = n-i-1; if(l<max_inversions){ max_inversions = l; } long long k2 = k; for(int j=min_inversions; j<=max_inversions; j++){ long long minus = get(n-i-1, l-j); //printf(" j=%d, minus=%lld\n", j, minus); long long k3 = k2 - minus; if(k3<0){ found = true; k = k2; l -= j; //printf("%dth\n", j); ans.push_back(nth(j)); break; } k2 = k3; } if(!found){ printf("NIE\n"); return 0; } } printf("TAK\n"); for(auto a: ans) printf("%d ", a); printf("\n"); return 0; }
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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 | #include <cstdio> #include <vector> #include <unordered_map> #include <set> using namespace std; #define INF 1000000000000000009LL struct pair_hash { template <class T1, class T2> std::size_t operator () (const std::pair<T1,T2> &p) const { auto h1 = std::hash<T1>{}(p.first); auto h2 = std::hash<T2>{}(p.second); return (h1 * 937ull) ^ h2; } }; unordered_map<pair<int, long long>, long long, pair_hash> c; // In how many ways we can get l inversions in n-permutation. long long get(long long n, long long l){ long long mx = n*(n-1)/2; long long half = mx/2; if (l>half) l = mx - l; if (l>200) return INF; if (l>20 && n>100) return INF; if (l>8 && n>1000) return INF; if (l==0) return 1; if (n==1) return 0; pair<int, long long> pp(n, l); auto it = c.find(pp); if (it != c.end()) return it->second; long long sum = 0; long long mm = n-1; if (l<mm) mm = l; for (long long i = 0; i <= mm; i++) { sum += get(n-1, l-i); if (sum > INF) { sum = INF; break; } } return c[pp] = sum; } set<int> remaining; void init(int n){ for(int i=0; i<n; i++){ remaining.insert(i+1); } } int nth(size_t n){ if(n>remaining.size()/2){ //if(0){ n = remaining.size() - n - 1; for(auto it = remaining.rbegin(); it!=remaining.rend(); it++){ if(n-- == 0){ int r = *it; remaining.erase(--it.base()); return r; } } } else{ for(auto it = remaining.begin(); it!=remaining.end(); it++){ if(n-- == 0){ int r = *it; remaining.erase(it); return r; } } } return -1; } void test(){ for(long long n=1; n<200000; n++){ int x=-1; for(int l=0; l<=n*(n-1)/2; l++){ x = l; if(get(n, l)==INF){ break; } } printf("%lld %d\n", n, x); } } int main(){ //test(); return 0; long long n; long long k; scanf("%lld %lld", &n, &k); k--; if((n*(n-1)) % 4){ printf("NIE\n"); return 0; } long long l = n*(n-1)/4; init(n); vector<int> ans; for(int i=0; i<n; i++){ //printf("i=%d, l=%lld, k=%lld\n", i, l, k); bool found=false; long long n1 = n-i-1; long long min_inversions = l - (n1*(n1-1))/2; if(min_inversions<0) min_inversions = 0; int max_inversions = n-i-1; if(l<max_inversions){ max_inversions = l; } long long k2 = k; for(int j=min_inversions; j<=max_inversions; j++){ long long minus = get(n-i-1, l-j); //printf(" j=%d, minus=%lld\n", j, minus); long long k3 = k2 - minus; if(k3<0){ found = true; k = k2; l -= j; //printf("%dth\n", j); ans.push_back(nth(j)); break; } k2 = k3; } if(!found){ printf("NIE\n"); return 0; } } printf("TAK\n"); for(auto a: ans) printf("%d ", a); printf("\n"); return 0; } |