This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
//one of the coolest problems i've ever seen
#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,e) for (ll i = s; i <= e; ++i)
#define pb(u) push_back(u)
#define len(a) (ll)a.size()
typedef long long ll;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
ll h, w, x = 0; cin >> h >> w;
ll a[h+1], b[w+1];
rep (i,1,h) cin >> a[i];
rep (i,1,w) cin >> b[i];
vector<ll> ha, hb;
ha.pb(1), hb.pb(1);
rep (i,2,h) {
while (true) {
ll n = len(ha);
if (n==1 || ((a[ha[n-1]]-a[ha[n-2]])*(i-ha[n-1])<(a[i]-a[ha[n-1]])*(ha[n-1]-ha[n-2]))) break;
ha.pop_back();
}
ha.pb(i);
}
rep (i,2,w) {
while (true) {
ll n = len(hb);
if (n==1 || ((b[hb[n-1]]-b[hb[n-2]])*(i-hb[n-1])<(b[i]-b[hb[n-1]])*(hb[n-1]-hb[n-2]))) break;
hb.pop_back();
}
hb.pb(i);
}
while (true) {
ll n = len(ha), m = len(hb);
if (max(n,m)<2) break;
if (n>1 && (m==1 || ((a[ha[n-1]]-a[ha[n-2]])*(hb[m-1]-hb[m-2])>(b[hb[m-1]]-b[hb[m-2]])*(ha[n-1]-ha[n-2])))) {
x += b[hb[m-1]] * (ha[n-1]-ha[n-2]);
ha.pop_back();
}
else {
x += a[ha[n-1]] * (hb[m-1]-hb[m-2]);
hb.pop_back();
}
}
cout << x << "\n";
return 0;
}
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