This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "homecoming.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;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define fbo find_by_order
#define ook order_of_key
#define INF 1e18
#define ret return
typedef long long ll;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector < pair<int, int> > vii;
typedef long double ld;
typedef tree<pair<int,int>, null_type, less<pair<int,int> >, rb_tree_tag, tree_order_statistics_node_update> pbds;
typedef set<int>::iterator sit;
typedef map<int,int>::iterator mit;
typedef vector<int>::iterator vit;
ll solve(int n, int k, int *A, int *B){
ll tA = 0;
ll tB = 0;
for(int i = 0; i < n; i++){
tA+=A[i];
tB+=B[i];
}
ll anss = max(ll(0),tA-tB);
if(k==1){
ll ans = 0;
for(int i = 0; i < n; i++){
ans+=max(0,A[i]-B[i]);
}
ret ans;
}
int N = n;
for(int T = 0; T < N; T++){
//cout << "**" << t << endl;
//banning the t-th book
vi a, b;
a.pb(0);
b.pb(0);
for(int i = 1; i < N; i++){
a.pb(A[(i+T)%N]);
b.pb(B[(i+T)%N]);
}
n=a.size();
ll dp[n][k+5];
ll cost[n];
cost[0]=b[0];
for(int i = 1; i < n; i++){
cost[i]=b[i]+cost[i-1];
}
for(int i = n-1; n-1<i+k-1; i--){
for(int j = k; j >= 0; j--){
if(i+j-1>n-1){
dp[i][j]=-INF;
}
else{
ll t=cost[i+j-1]-cost[i-1];
dp[i][j]=max(dp[i][j+1],-t);
}
}
}
for(int i = n-k; i >= 1; i--){
for(int j = k; j >= 0; j--){
if(j==k){
dp[i][j]=A[i]-B[i]+dp[i+1][j-1];
}
else{
ll t=cost[i+j-1]-cost[i-1];
dp[i][j]=max(dp[i][j+1],-t+dp[i+j][0]);
}
}
if(dp[i][0]<0)while(1){};
}
anss=max(anss,dp[1][0]);
}
ret anss;
}
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