# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
139513 | eriksuenderhauf | 이상한 기계 (APIO19_strange_device) | C++11 | 3006 ms | 17432 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#define mem(a,v) memset((a), (v), sizeof (a))
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%lld", &(n))
#define nai(a, n) for (int i = 0; i < (n); i++) ni(a[i])
#define nal(a, n) for (int i = 0; i < (n); i++) nl(a[i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%lld\n", (n))
#define pii pair<int, int>
#define pil pair<int, long long>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vil vector<pil>
#define vll vector<pll>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef cc_hash_table<int,int,hash<int>> ht;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> oset;
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const ll INF = 1e18;
const int MAXN = 1e6 + 5;
const double eps = 1e-9;
int main() {
int n; ll a, b; scanf("%d %I64d %I64d", &n, &a, &b);
ll k = a / __gcd(a,b+1), mx = 1, ans = 0;
if (k > (INF+b-1) / b) mx = INF+1;
else mx = k*b;
vll act;
for (int i = 0; i < n; i++) {
ll l, r; scanf("%I64d %I64d", &l, &r);
bool fl = (r-l+1 >= mx);
if (fl) {
act.pb(mp(0,mx-1));
break;
}
l %= mx, r %= mx;
if (l > r) {
act.pb(mp(l,mx-1));
act.pb(mp(0,r));
} else {
act.pb(mp(l,r));
}
}
n = act.size();
sort(act.begin(), act.end());
for (int i = 0; i < n; i++) {
int j = i;
ll curmx = act[i].se;
for (; j < n; j++) {
if (curmx < act[j].fi)
break;
curmx = max(curmx, act[j].se);
}
ans += curmx - act[i].fi + 1;
i = j-1;
}
prl(ans);
return 0;
}
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