이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
using namespace std;
typedef long long int lld;
#define INF 10000000000000000
lld DP[1000000][2];
int n,k;
lld C[1000000];
lld D[1000000];
lld compute(int pos, int type){
if(DP[pos][type]!=-1)return DP[pos][type];
if(type==0){
DP[pos][type]=max(compute(pos-1,0),compute(pos-1,1)+D[pos-1]);
}
if(type==1){
DP[pos][type]=max(compute(pos-1,0),compute(pos-1,1)+C[pos-1]);
}
return DP[pos][type];
}
lld solveLin(int A[], int B[]){
for(int i=0;i<n;i++){
C[i]=A[i]-B[i];
}
D[0]=A[0];
for(int i=0;i<k;i++)D[0]-=B[i];
for(int i=0;i+k<n;i++){
D[i+1]=D[i]-A[i]+B[i]+A[i+1]-B[i+k];
}
for(int i=n-k+1;i<n;i++){
D[i]=-INF;
}
for(int i=0;i<=n;i++){
DP[i][0]=-1;
DP[i][1]=-1;
}
DP[0][0]=0;
DP[0][1]=0;
//cout<<compute(n-1,0)<<" "<<compute(n-1,1)<<" "<<compute(n-1,2)<<endl;
/*for(int i=0;i<=n;i++){
for(int j=0;j<2;j++)cout<<compute(i,j)<<" ";
cout<<endl;
}*/
return compute(n,0);
}
long long solve(int N, int K, int *A, int *B){
n=N;
k=K;
lld sol=0;
for(int i=0;i<n;i++)sol+=A[i]-B[i];
sol=max(sol,(lld)0);
for(int i=0;i<n;i++){
sol=max(sol,solveLin(A,B));
for(int j=0;j<n-1;j++){
swap(A[j],A[j+1]);
swap(B[j],B[j+1]);
}//for(int j=0;j<n;j++)cout<<A[j]<<" "<<B[j]<<endl;
}
return sol;
}
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