| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 237652 | zaneyu | Gap (APIO16_gap) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*input
2 4
2 3 6 8
*/
//#include "gap.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
#define ll long long
#define ld long double
#define pii pair<double,double>
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e12;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
ll mult(ll a,ll b){
return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
if(b<=0) return 1;
ll res=1LL;
while(b){
if(b&1){
res=(res*a)%MOD;
}
a=(a*a)%MOD;
b>>=1;
}
return res;
}
const ll maxn=2e5+5;
const ll maxlg=__lg(maxn)+2;
static void my_assert(int k){ if (!k) exit(1); }
static int subtask_num, N;
static long long A[100001];
static long long call_count;
void MinMax(long long s, long long t, long long *mn, long long *mx)
{
int lo = 1, hi = N, left = N+1, right = 0;
my_assert(s <= t && mn != NULL && mx != NULL);
while (lo <= hi){
int mid = (lo+hi)>>1;
if (A[mid] >= s) hi = mid - 1, left = mid;
else lo = mid + 1;
}
lo = 1, hi = N;
while (lo <= hi){
int mid = (lo+hi)>>1;
if (A[mid] <= t) lo = mid + 1, right = mid;
else hi = mid - 1;
}
if (left > right) *mn = *mx = -1;
else{
*mn = A[left];
*mx = A[right];
}
if (subtask_num == 1) call_count++;
else if (subtask_num == 2) call_count += right-left+2;
}
ll arr[maxn];
long long findGap(int T, int n){
if(T==1){
int l=0,r=n-1;
MinMax(0,(ll)1e18,arr+l,arr+r);
++l;
--r;
while(l<=r){
MinMax(arr[l-1]+1,arr[r+1]-1,arr+l,arr+r);
++l;
--r;
}
ll ans=0;
REP1(i,n-1) MXTO(ans,arr[i]-arr[i-1]);
return ans;
}
else{
//pigeon
ll l,r;
MinMax(0,(ll)1e18,&l,&r);
ll sz=(r-l)/(n-1);
vector<ll> v;
v.pb(l);
REP(i,n-1){
ll tl,tr;
MinMax(l+i*sz,l+(i+1)*sz-1,&tl,&tr);
if(tl!=-1){
v.pb(tl);
v.pb(tr);
}
}
v.pb(r);
ll ans=0;
REP1(i,sz(v)-1){
MXTO(ans,v[i]-v[i-1]);
}
return ans;
}
}
int main()
{
FILE *in = stdin, *out = stdout;
my_assert(2 == fscanf(in, "%d%d", &subtask_num, &N));
my_assert(1 <= subtask_num && subtask_num <= 2);
my_assert(2 <= N && N <= 100000);
for (int i=1;i<=N;i++) my_assert(1 == fscanf(in, "%lld", A+i));
for (int i=1;i<N;i++) my_assert(A[i] < A[i+1]);
fprintf(out, "%lld\n", findGap(subtask_num, N));
fprintf(out, "%lld\n", call_count);
}