이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "ricehub.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
long long ma=-1;
int besthub(int R, int L, int X[], long long B){
/* if(R<=500){ ma=-1;
for(int i=0;i<R;i++){
for(int j=i;j<R;j++){
// cout<<i<<" "<<j<<endl;
vector<int>rp;
for(int k=i;k<=j;k++){
rp.push_back(X[k]);
}
int mi=(i+j)/2;
mi=X[mi];
// cout<<i<<" "<<j<<endl;
bool sw=0;
long long b=B;
for(int k=0;k<rp.size();k++){
b-=abs(mi-rp[k]);
if(b<0){
sw=1;
break;
}
//c++;
}
if(sw==0){
// cout<<i<<" "<<j<<" "<<rp.size()<<endl;
long long tam=rp.size();
ma=max(ma,tam);
}
}
}
return ma;
}
else{*/
vector<ll>T;
T.push_back(0);
for(int i=0;i<R;i++){
T.push_back(X[i]+T[T.size()-1]);
}
for(int i=0;i<R;i++){
int l,r;
l=i+1;
r=R;
while(true){
ll mi=l+(r-l)/2;
//cout<<l<<" "<<r<<" "<<ma<<" "<<mi<<endl;
if(l>r) break;
ll p=(i+mi)/2;
if((p-i)*X[p]-(T[p]-T[i])+(T[mi+1]-T[p+1])-(mi-p)*X[p]<=B){
ma=max(ma,(ll)(mi-i)+1);
if(l==mi) break;
l=mi;
}
else{
if(r==mi) break;
r=mi;
}
}
}
return ma;
//}
}
/*
#define MAX_R 1000000
static int R, L;
static long long B;
static int X[MAX_R];
static int solution;
inline
void my_assert(int e) {if (!e) abort();}
static void read_input()
{
int i;
my_assert(3==scanf("%d %d %lld",&R,&L,&B));
for(i=0; i<R; i++)
my_assert(1==scanf("%d",&X[i]));
my_assert(1==scanf("%d",&solution));
}
int main()
{
int ans;
read_input();
ans = besthub(R,L,X,B);
if(ans==solution)
printf("Correct.\n");
else
printf("Incorrect. Returned %d instead of %d.\n",ans,solution);
return 0;
}
*/
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