Submission #213016

#	Time	Username	Problem	Language	Result	Execution time	Memory
213016		arman_ferdous	Strange Device (APIO19_strange_device)	C++17	10 / 100	607 ms	53320 KiB

strange_device

/* 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 ll oo = 2e18;

vector<pair<ll,ll>> interval;

int main () {
  int n;
  ll A, B;
  scanf("%d %lld %lld", &n, &A, &B);

  int inf = 0, all = 0; ll sum = 0;
  ll g = __gcd(A, B + 1);
  ll K = A / g;
  ll T = K * B;

  
  if(K > oo / B) inf = 1;
  assert(inf == 0);
  while(n--) {
    ll l, r;
    scanf("%lld %lld", &l, &r);

    if(r - l + 1 >= T) {
      all = 1;
      break;
    } 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 if(all) printf("%lld\n", T);
  else {
    ll ans = 0;
    sort(interval.begin(), interval.end());
    for(int i = 0; i < (int)interval.size(); i++) {
      ans += interval[i].second - interval[i].first + 1;
      if(i) ans -= max(interval[i - 1].second - interval[i].first + 1, 0ll);
    }
    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:33:10: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
     scanf("%lld %lld", &l, &r);
     ~~~~~^~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 0 / 10
• Subtask 2: 0 / 5
• Subtask 3: 0 / 5
• Subtask 4: 5 / 5
• Subtask 5: 5 / 5
• Subtask 6: 0 / 20
• Subtask 7: 0 / 20
• Subtask 8: 0 / 30