Submission #679031

#TimeUsernameProblemLanguageResultExecution timeMemory
679031sunwukong123Cigle (COI21_cigle)C++14
0 / 100
3 ms5332 KiB
#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 dp[MAXN][MAXN]; int mn[MAXN][MAXN]; int n; int A[MAXN]; int ans; int32_t main() { cin >> n; FOR(i,1,n) cin >> A[i]; FOR(i,0,n)mn[i][0]=mn[0][i]=-inf; FOR(i,1,n) { int sum = 0, same=0, lf = 0, rg = 0, idx=i; FOR(j,i,n) { // count the equal for the previous guy here. dp[i][j]=dp[i][j-1]; // this guy has nothing underneath chmax(dp[i][j], dp[idx][i-1] + sum - same); // the last equal can't be counted if the previous row doesn't have a thing after this chmax(dp[i][j], mn[idx-1][i-1] + sum); // the bottom row has 1 more than this. rg += A[j]; while(rg>lf && idx>1) { --idx; lf += A[idx]; } sum += (lf == rg); if(lf>=rg)same=(lf==rg); // there's } chmax(ans, dp[i][n]); } cout << ans; }
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