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;
#define int long long
#define FOR(i,s,e) for(int i = s; i <= (int)e; ++i)
#define DEC(i,s,e) for(int i = s; i >= (int)e; --i)
#define IAMSPEED ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef LOCAL
#define db(x) cerr << #x << "=" << x << "\n"
#define db2(x, y) cerr << #x << "=" << x << " , " << #y << "=" << y << "\n"
#define db3(a,b,c) cerr<<#a<<"="<<a<<","<<#b<<"="<<b<<","<<#c<<"="<<c<<"\n"
#define dbv(v) cerr << #v << ":"; for (auto ite : v) cerr << ite << ' '; cerr <<"\n"
#define dbvp(v) cerr << #v << ":"; for (auto ite : v) cerr << "{" << ite.f << ',' << ite.s << "} "; cerr << "\n"
#define dba(a,ss,ee) cerr << #a << ":"; FOR(ite,ss,ee) cerr << a[ite] << ' '; cerr << "\n"
#define reach cerr << "LINE: " << __LINE__ << "\n";
#else
#define reach
#define db(x)
#define db2(x,y)
#define db3(a,b,c)
#define dbv(v)
#define dbvp(v)
#define dba(a,ss,ee)
#endif
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define f first
#define s second
#define g0(x) get<0>(x)
#define g1(x) get<1>(x)
#define g2(x) get<2>(x)
#define g3(x) get<3>(x)
typedef pair <int, int> pi;
typedef tuple<int,int,int> ti3;
typedef tuple<int,int,int,int> ti4;
int rand(int a, int b) { return a + rng() % (b-a+1); }
const int MOD = 1e9 + 7;
const int inf = (int)1e9 + 500;
const long long oo = (long long)1e18 + 500;
template <typename T> bool chmax(T& a, const T b) { return a<b ? a = b, 1 : 0; }
template <typename T> bool chmin(T& a, const T b) { return a>b ? a = b, 1 : 0; }
const int MAXN = 5005;
int n;
int A[MAXN];
int dp[MAXN][MAXN];
int g[MAXN][MAXN];
int ans = 0;
int32_t main() {
cin >> n;
FOR(i,1,n) cin >> A[i];
FOR(i,0,n)g[0][i]=g[i][0]=-inf;
FOR(i,1,n) {
int idx=i;
bool same=0;
int sum =0;
int lf=0,rg=0;
FOR(j,i,n) {
dp[i][j]=max({dp[i][j-1], dp[idx][i-1] + sum - same, g[idx-1][i-1] + sum});
g[i][j]=max(g[i-1][j],dp[i][j]);
/* either
just add this to this existing row. don't pair this guy with anything below
pair this guy with something below. cost=no. of places equal, except the last one.
pair this guy with something below. the row below exceeds my row, so there's basically no more pairing after this point.
*/
rg+=A[j];
while(idx>1 && rg>lf){
--idx;
lf+=A[idx];
}
sum += (lf == rg);
if(lf>=rg)same = (lf==rg);
}
chmax(ans,dp[i][n]);
}
cout << ans;
}
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