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 se second
#define fs first
#define pb push_back
#define ll long long
#define ii pair<ll,ll>
#define ld long double
#define SZ(v) (int)v.size()
#define ALL(v) v.begin(), v.end()
#define bit(msk, i) ((msk >> i) & 1)
#define iter(id, v) for(auto id : v)
#define rep(i,m,n) for(int i=(m); i<=(n); i++)
#define reb(i,m,n) for(int i=(m); i>=(n); i--)
using namespace std;
mt19937_64 rd(chrono :: steady_clock :: now().time_since_epoch().count());
ll Rand(ll l, ll r) { return uniform_int_distribution<ll> (l, r)(rd); }
const int N = 5e3 + 7;
const ll Mod = 1e9 + 7;
const int szBL = 916;
const ll INF = 1e9;
const int BASE = 137;
int n;
int a[N];
int dp[N][N], f[N];
void solution () {
cin >> n;
rep (i, 1, n) cin >> a[i];
rep (L, 1, n) {
f[0] = -INF;
rep (i, 1, L - 1) {
f[i] = max(f[i - 1], dp[i][L - 1]);
}
int ptr = L - 1, curcost = 0, suf = a[L - 1], pre = a[L];
dp[L][L - 1] = f[L - 1];
rep (R, L, n) {
dp[L][R] = max(dp[L][R], dp[L][R - 1]);
while (ptr > 1 && suf < pre) {
dp[L][R] = max (dp[L][R], dp[ptr - 1][L - 1] + curcost);
suf += a[ptr - 1];
--ptr;
}
if (suf == pre) {
curcost++;
dp[L][R + 1] = max (dp[L][R + 1], f[ptr - 1] + curcost);
}
// cout << L<<","<<R<<" "<<dp[L][R] <<" "<<suf<<","<<pre<<": "<<f[ptr - 1] + curcost<<"\n";
pre += a[R + 1];
}
}
int res = 0;
rep (L, 1, n) {
res = max(res, dp[L][n]);
}
cout << res <<"\n";
}
#define file(name) freopen(name".inp", "r", stdin); freopen(name".out", "w", stdout);
int main () {
// file("c");
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
ll num_Test = 1;
// cin >> num_Test;
while(num_Test--)
solution();
}
/*
18 3
2 5
6 21
13 19
9 17
14 17
19 20
2 16
2 10
9 14
19 20
14 16
1 3
17 19
14 21
18 19
4 7
5 12
1 13
*/
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