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>
#define fr first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;
template<class T>void umax(T &a,T b){if(a<b)a=b;}
template<class T>void umin(T &a,T b){if(b<a)a=b;}
#ifdef juggernaut
#define printl(args...) printf(args)
#else
#define printl(args...) 0
#endif
ll solve2(int n,int k,deque<ll>A,deque<ll>B){
vector<ll>a,b,pref,dp,cost;
b=cost=a=pref={0};
for(int i=0;i<2*n;i++){
a.push_back(A[i%n]);
pref.push_back(pref.back()+a.back());
b.push_back(B[i%n]);
cost.push_back(cost.back()+b.back());
}
dp.assign(2*n+1,0);
for(int i=n+1;i<=2*n;i++){
dp[i]=dp[i-1];
for(int j=n+1;j<=i;j++)if(i-j+1>=k)
umax(dp[i],dp[j-1]+(pref[i+1-k]-pref[j-1])-(cost[i]-cost[j-1]));
}
return max(dp.back(),accumulate(A.begin(),A.end(),0ll)-accumulate(B.begin(),B.end(),0ll));
}
ll solve(int n,int k,int *A,int *B){
int N=n+2;
ll ans=0;
deque<ll>x,y;
for(int i=0;i<n;i++){
x.push_back(A[i]);
y.push_back(B[i]);
}
while(N--){
umax(ans,solve2(n,k,x,y));
x.push_back(x.front());
x.pop_front();
y.push_back(y.front());
y.pop_front();
}
return ans;
}
#ifdef juggernaut
int main() {
int T; assert(scanf("%d", &T) == 1);
for(int t = 0; t < T; t++) {
int N, K; assert(scanf("%d%d", &N, &K) == 2);
int *A = new int[N];
int *B = new int[N];
for(int i = 0; i < N; i++) assert(scanf("%d", &A[i]) == 1);
for(int i = 0; i < N; i++) assert(scanf("%d", &B[i]) == 1);
printf("%lld\n", solve(N, K, A, B));
delete[] A;
delete[] B;
}
return 0;
}
#endif
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