// Author : Jakub Rożek // Task : Bardzo ulubiony ciąg // Contest : PA 2024 r5 C // Memory : O(n^3) // Time : O(n^3) // Solution : Rozwiązanie poeinno dzialac #include "bits/stdc++.h" using namespace std; using LL = long long; template <typename T> using P = pair<T, T>; template <typename T> using VV = vector<vector<T>>; #define FOR(i,a,b) for(int i=(a); i<=(b); ++i) #define FORD(i,a,b) for(int i=(a); i>=(b); --i) #define REP(i,n) for(int i=0; i<(n); ++i) #define all(x) x.begin(), x.end() #define ssize(x) int((x).size()) #ifdef DEBUG template <typename T1, typename T2> auto&operator<<(auto&o,pair<T1,T2>p){return o<<'('<<p.first<<", "<<p.second<<")";} auto operator<<(auto&o,auto x)->decltype(x.end(),o){o<<"{";for(auto e:x)o<<","<<e;return o<<"}";} #define debug(x...) cerr<<"["#x"]: ",[](auto...$){((cerr<<$<<"; "),...)<<endl;}(x) #else #define debug(...) {} #endif // const LL INF = 1'000'000'000'000'000'018; // const int INF = 1'000'000'009; // const LL mod = 1'000'000'007; const int N = 500; struct trzy { int a,b,c; }; int n, x, y, z, w; LL odp, sum; LL sum_pref[N+1]; int dp[N+1][N+1][N+1]; unordered_map <LL, LL> ile; unordered_map <LL, vector<trzy>> zbior; void skaluj(auto &v) { map <int, int> skala; set <int> wartosci; skala.clear(); wartosci.clear(); for (auto i : v) wartosci.insert(i.a); x = 0; for (auto i : wartosci) { ++x; skala[i] = x; } for (auto &i : v) i.a = skala[i.a]; skala.clear(); wartosci.clear(); for (auto i : v) wartosci.insert(i.b); y = 0; for (auto i : wartosci) { ++y; skala[i] = y; } for (auto &i : v) i.b = skala[i.b]; skala.clear(); wartosci.clear(); for (auto i : v) wartosci.insert(i.c); z = 0; for (auto i : wartosci) { ++z; skala[i] = z; } for (auto &i : v) i.c = skala[i.c]; } void solution() { cin >> n; FOR (i, 1, n) { cin >> sum_pref[i]; sum_pref[i] += sum_pref[i-1]; } FOR (a, 0, n) { FOR (b, 0, n) { FOR (c, 0, n) { zbior[sum_pref[a]+sum_pref[b]+sum_pref[c]].push_back({a,b,c}); } } } // dodaje wszystkie 3 for (auto [_, v] : zbior) { skaluj(v); FOR (a, 0, x) { FOR (b, 0, y) { FOR (c, 0, z) { dp[a][b][c] = 0; } } } for (auto i : v) { dp[i.a][i.b][i.c] += 1; } FOR (a, 1, x) { FOR (b, 1, y) { FOR (c, 1, z) { if (dp[a][b][c]) odp += dp[a-1][b-1][c-1]; dp[a][b][c] += dp[a-1][b][c]; dp[a][b][c] += dp[a][b-1][c]; dp[a][b][c] += dp[a][b][c-1]; dp[a][b][c] -= dp[a-1][b-1][c]; dp[a][b][c] -= dp[a][b-1][c-1]; dp[a][b][c] -= dp[a-1][b][c-1]; dp[a][b][c] += dp[a-1][b-1][c-1]; } } } } // odejmuje 2 takie same FOR (a, 0, n-1) { FOR (b, a+1, n) { ile[sum_pref[b]-sum_pref[a]]++; } } x = 0; FOR (a, 0, n-1) { FOR (b, a+1, n) { ++x; odp -= 3 * ile[-(sum_pref[b]-sum_pref[a])*2]; } } // dodaje spowrotem 3 takie same FOR (a, 0, n-1) { FOR (b, a+1, n) { if (sum_pref[b]-sum_pref[a] == 0) odp += 2; } } cout << odp / 6 << '\n'; return; } int main() { cin.tie(0)->sync_with_stdio(0); int tests = 1; // cin>>tests; FOR (i, 1, tests) { solution(); } 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 | // Author : Jakub Rożek // Task : Bardzo ulubiony ciąg // Contest : PA 2024 r5 C // Memory : O(n^3) // Time : O(n^3) // Solution : Rozwiązanie poeinno dzialac #include "bits/stdc++.h" using namespace std; using LL = long long; template <typename T> using P = pair<T, T>; template <typename T> using VV = vector<vector<T>>; #define FOR(i,a,b) for(int i=(a); i<=(b); ++i) #define FORD(i,a,b) for(int i=(a); i>=(b); --i) #define REP(i,n) for(int i=0; i<(n); ++i) #define all(x) x.begin(), x.end() #define ssize(x) int((x).size()) #ifdef DEBUG template <typename T1, typename T2> auto&operator<<(auto&o,pair<T1,T2>p){return o<<'('<<p.first<<", "<<p.second<<")";} auto operator<<(auto&o,auto x)->decltype(x.end(),o){o<<"{";for(auto e:x)o<<","<<e;return o<<"}";} #define debug(x...) cerr<<"["#x"]: ",[](auto...$){((cerr<<$<<"; "),...)<<endl;}(x) #else #define debug(...) {} #endif // const LL INF = 1'000'000'000'000'000'018; // const int INF = 1'000'000'009; // const LL mod = 1'000'000'007; const int N = 500; struct trzy { int a,b,c; }; int n, x, y, z, w; LL odp, sum; LL sum_pref[N+1]; int dp[N+1][N+1][N+1]; unordered_map <LL, LL> ile; unordered_map <LL, vector<trzy>> zbior; void skaluj(auto &v) { map <int, int> skala; set <int> wartosci; skala.clear(); wartosci.clear(); for (auto i : v) wartosci.insert(i.a); x = 0; for (auto i : wartosci) { ++x; skala[i] = x; } for (auto &i : v) i.a = skala[i.a]; skala.clear(); wartosci.clear(); for (auto i : v) wartosci.insert(i.b); y = 0; for (auto i : wartosci) { ++y; skala[i] = y; } for (auto &i : v) i.b = skala[i.b]; skala.clear(); wartosci.clear(); for (auto i : v) wartosci.insert(i.c); z = 0; for (auto i : wartosci) { ++z; skala[i] = z; } for (auto &i : v) i.c = skala[i.c]; } void solution() { cin >> n; FOR (i, 1, n) { cin >> sum_pref[i]; sum_pref[i] += sum_pref[i-1]; } FOR (a, 0, n) { FOR (b, 0, n) { FOR (c, 0, n) { zbior[sum_pref[a]+sum_pref[b]+sum_pref[c]].push_back({a,b,c}); } } } // dodaje wszystkie 3 for (auto [_, v] : zbior) { skaluj(v); FOR (a, 0, x) { FOR (b, 0, y) { FOR (c, 0, z) { dp[a][b][c] = 0; } } } for (auto i : v) { dp[i.a][i.b][i.c] += 1; } FOR (a, 1, x) { FOR (b, 1, y) { FOR (c, 1, z) { if (dp[a][b][c]) odp += dp[a-1][b-1][c-1]; dp[a][b][c] += dp[a-1][b][c]; dp[a][b][c] += dp[a][b-1][c]; dp[a][b][c] += dp[a][b][c-1]; dp[a][b][c] -= dp[a-1][b-1][c]; dp[a][b][c] -= dp[a][b-1][c-1]; dp[a][b][c] -= dp[a-1][b][c-1]; dp[a][b][c] += dp[a-1][b-1][c-1]; } } } } // odejmuje 2 takie same FOR (a, 0, n-1) { FOR (b, a+1, n) { ile[sum_pref[b]-sum_pref[a]]++; } } x = 0; FOR (a, 0, n-1) { FOR (b, a+1, n) { ++x; odp -= 3 * ile[-(sum_pref[b]-sum_pref[a])*2]; } } // dodaje spowrotem 3 takie same FOR (a, 0, n-1) { FOR (b, a+1, n) { if (sum_pref[b]-sum_pref[a] == 0) odp += 2; } } cout << odp / 6 << '\n'; return; } int main() { cin.tie(0)->sync_with_stdio(0); int tests = 1; // cin>>tests; FOR (i, 1, tests) { solution(); } return 0; } |