Submission #935582

#TimeUsernameProblemLanguageResultExecution timeMemory
935582sysiaStrange Device (APIO19_strange_device)C++17
100 / 100
985 ms163008 KiB
//Sylwia Sapkowska #include <bits/stdc++.h> #pragma GCC optimize("O3", "unroll-loops") using namespace std; void __print(int x) {cerr << x;} void __print(long long x) {cerr << x;} void __print(long double x) {cerr << x;} void __print(char x) {cerr << "'" << x << "'";} void __print(const char *x) {cerr << '"' << x << '"';} void __print(const string &x) {cerr << '"' << x << '"';} void __print(bool x) {cerr << (x ? "true" : "false");} template<typename T, typename V> void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';} template<typename T> void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";} void _print() {cerr << "]\n";} template <typename T, typename... V> void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);} #ifdef LOCAL #define debug(x...) cerr << "[" << #x << "] = ["; _print(x) #else #define debug(x...) #endif #define int long long typedef pair<int, int> T; const int oo = 1e18, oo2 = 1e9+7, K = 30; const int mod = 998244353; #define ii __int128_t void solve(){ int n, A, B; cin >> n >> A >> B; //po obrzydliwych obliczeniach dostajemy ze co AB to sie powtarza + dziala na samplach //jesli s \neq t mod AB, to daja rozne pary, nie wprost daja //wiec musi s \equiv t mod B, ale wtedy s \neq t mod A (bo gdyby zachodzilo przystawanie, to mozna wymnozyc(?)) //czy to ze AB niekoniecznie wzglednie pierwsze przeszkadza w czyms? //wiec wtedy pierwsza wartosc daje inna reszte set<T>s;//przedzialy mod AB ii CC = (ii)A*(ii)B/(ii)__gcd(A, B+1); CC = min(CC, (ii)2*oo); int C = CC; for (int i = 0; i<n; i++){ int l, r; cin >> l >> r; int L = l%C; int R = r%C; if (r-l+1 >= C) { s.insert({0, C-1}); } else { if (L <= R) s.insert({L, R}); else { s.insert({L, C-1}); s.insert({0, R}); } } } debug(s); map<int, int>sweep; for (auto [a, b]: s){ sweep[a]++; sweep[b+1]--; } int prev = -1; int open = 0, ans = 0; for (auto [a, val]: sweep){ debug(a, val, open); if (open) { ans += a-prev; } open += val; prev = a; } cout << ans << "\n"; } int32_t main(){ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t = 1; //cin >> t; while (t--) solve(); return 0; }
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