Submission #213053

#TimeUsernameProblemLanguageResultExecution timeMemory
213053arman_ferdousStrange Device (APIO19_strange_device)C++17
100 / 100
589 ms53316 KiB
/* at time = 0, (x, y) = (0, 0) * we need to find minimum t > 0, such that (x, y) = 0 again ie. the cycle period. * now, y = 0 = t % B => t = K * B (for some integer K) * putting t in x's eqn and simplifying gives us a nice formula * K * (B + 1) % A = 0 * from here it is easy to find K's value. * answering queries are just merging intervals */ #include <bits/stdc++.h> using namespace std; using ll = long long; const double oo = 1e19; vector<pair<ll,ll>> interval; int main () { int n; ll A, B; scanf("%d %lld %lld", &n, &A, &B); int inf = 0; ll sum = 0; ll g = __gcd(A, B + 1); ll K = A / g; ll T = K * B; if(K * 1.0 > oo / B * 1.0) inf = 1; while(n--) { ll l, r; scanf("%lld %lld", &l, &r); if(!inf && r - l + 1 >= T) { printf("%lld\n", T); return 0; } sum += r - l + 1; ll le = l % T; ll ri = r % T; if(le <= ri) { interval.push_back({le, ri}); } else { interval.push_back({0, ri}); interval.push_back({le, T - 1}); } } if(inf) printf("%lld\n", sum); else { ll ans = 0, ptr = -1; sort(interval.begin(), interval.end()); for(int i = 0; i < (int)interval.size(); i++) { if(ptr < interval[i].first) { ans += interval[i].second - interval[i].first + 1; ptr = interval[i].second; } else { ans += max(0ll, interval[i].second - ptr); ptr = max(ptr, interval[i].second); } } printf("%lld\n", ans); } return 0; }

Compilation message (stderr)

strange_device.cpp: In function 'int main()':
strange_device.cpp:21:8: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   scanf("%d %lld %lld", &n, &A, &B);
   ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~
strange_device.cpp:31:10: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
     scanf("%lld %lld", &l, &r);
     ~~~~~^~~~~~~~~~~~~~~~~~~~~
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