結果
問題 | No.1117 数列分割 |
ユーザー |
|
提出日時 | 2020-07-17 22:23:16 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
TLE
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 6,965 bytes |
コンパイル時間 | 1,784 ms |
コンパイル使用メモリ | 131,000 KB |
実行使用メモリ | 104,584 KB |
最終ジャッジ日時 | 2024-11-30 03:27:29 |
合計ジャッジ時間 | 26,408 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 23 TLE * 3 |
ソースコード
#pragma GCC target ("avx")#pragma GCC optimize("Ofast")#pragma GCC optimize("unroll-loops")//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")#define _USE_MATH_DEFINES#include<iostream>#include<string>#include<queue>#include<cmath>#include<map>#include<set>#include<list>#include<iomanip>#include<vector>#include<random>#include<functional>#include<algorithm>#include<stack>#include<cstdio>#include<cstring>#include<bitset>#include<unordered_map>#include<climits>#include<fstream>#include<complex>#include<time.h>#include<cassert>#include<functional>#include<numeric>#include<tuple>using namespace std;using ll = long long;using ld = long double;#define int long long#define all(a) (a).begin(),(a).end()#define fs first#define sc second#define xx first#define yy second.first#define zz second.second#define H pair<int, int>#define P pair<int, pair<int, int>>#define Q(i,j,k) mkp(i,mkp(j,k))#define rng(i,s,n) for(int i = (s) ; i < (n) ; i++)#define rep(i,n) rng(i, 0, (n))#define mkp make_pair#define vec vector#define vi vec<int>#define pb emplace_back#define siz(a) (int)(a).size()#define crdcomp(b) sort(all((b)));(b).erase(unique(all((b))),(b).end())#define getidx(b,i) (lower_bound(all(b),(i))-(b).begin())#define ssp(i,n) (i==(int)(n)-1?"\n":" ")#define ctoi(c) (int)(c-'0')#define itoc(c) (char)(c+'0')#define cyes printf("Yes\n")#define cno printf("No\n")#define cdf(n) int quetimes_=(n);rep(qq123_,quetimes_)#define gcj printf("Case #%lld: ",qq123_+1)#define readv(a,n) a.resize(n,0);rep(i,(n)) a[i]=read()#define found(a,x) (a.find(x)!=a.end())//#define endl "\n"constexpr int mod = (ll)1e9 + 7;constexpr int Mod = 998244353;constexpr ld EPS = 1e-10;constexpr ll inf = (ll)3 * 1e18;constexpr int Inf = (ll)15 * 1e8;constexpr int dx[] = { -1,1,0,0 }, dy[] = { 0,0,-1,1 };template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }ll read() { ll u, k = scanf("%lld", &u); return u; }string reads() { string s; cin >> s; return s; }H readh(bool g = 0) { H u; int k = scanf("%lld %lld", &u.fs, &u.sc); if (g) u.fs--, u.sc--; return u; }bool ina(H t, int h, int w) { return 0 <= t.fs && t.fs < h && 0 <= t.sc && t.sc < w; }bool ina(int t, int l, int r) { return l <= t && t < r; }ll gcd(ll i, ll j) { return j ? gcd(j, i % j) : i; }ll popcount(ll x) {int sum = 0; for (int i = 0; i < 60; i++)if ((1ll << i) & x) sum++;return sum;}class mint {public:ll v;mint(ll v = 0) { s(v % mod + mod); }constexpr static int mod = 1e9 + 7;constexpr static int fn_ = 3e5 + 5;static mint fact[fn_], comp[fn_];mint pow(int x) const {mint b(v), c(1);while (x) {if (x & 1) c *= b;b *= b;x >>= 1;}return c;}inline mint& s(int vv) {v = vv < mod ? vv : vv - mod;return *this;}inline mint inv()const { return pow(mod - 2); }inline mint operator-()const { return mint() - *this; }inline mint& operator+=(const mint b) { return s(v + b.v); }inline mint& operator-=(const mint b) { return s(v + mod - b.v); }inline mint& operator*=(const mint b) { v = v * b.v % mod; return *this; }inline mint& operator/=(const mint b) { v = v * b.inv().v % mod; return *this; }inline mint operator+(const mint b) const { return mint(v) += b; }inline mint operator-(const mint b) const { return mint(v) -= b; }inline mint operator*(const mint b) const { return mint(v) *= b; }inline mint operator/(const mint b) const { return mint(v) /= b; }friend ostream& operator<<(ostream& os, const mint& m) {return os << m.v;}friend istream& operator>>(istream& is, mint& m) {int x; is >> x; m = mint(x);return is;}bool operator<(const mint& r)const { return v < r.v; }bool operator>(const mint& r)const { return v > r.v; }bool operator<=(const mint& r)const { return v <= r.v; }bool operator>=(const mint& r)const { return v >= r.v; }bool operator==(const mint& r)const { return v == r.v; }bool operator!=(const mint& r)const { return v != r.v; }explicit operator bool()const { return v; }explicit operator int()const { return v; }mint comb(mint k) {if (k > * this) return mint();if (!fact[0]) combinit();if (v >= fn_) {if (k > * this - k) k = *this - k;mint tmp(1);for (int i = v; i >= v - k.v + 1; i--) tmp *= mint(i);return tmp * comp[k.v];}return fact[v] * comp[k.v] * comp[v - k.v];}//nCkmint perm(mint k) {if (k > * this) return mint();if (!fact[0]) combinit();if (v >= fn_) {mint tmp(1);for (int i = v; i >= v - k.v + 1; i--) tmp *= mint(i);return tmp;}return fact[v] * comp[v - k.v];}static void combinit() {fact[0] = 1;for (int i = 1; i < fn_; i++) fact[i] = fact[i - 1] * mint(i);comp[fn_ - 1] = fact[fn_ - 1].inv();for (int i = fn_ - 2; i >= 0; i--) comp[i] = comp[i + 1] * mint(i + 1);}}; mint mint::fact[fn_], mint::comp[fn_];//--------------------------------------------------------------//---------------------------------------------------------------------class BIT {int size;int dat[500000];public:BIT() {}void init(int n) {size = n;for (int i = 1; i <= n; i++) dat[i] = 0;}void add(int i, int x) {i++;while (i <= size) {dat[i] += x;i += i & -i;}}//0-indexedvoid add(int l, int r, int x) {add(l, x); add(r, -x);}//[l,r)int query(int i) {i++;int sum = 0;while (i > 0) {sum += dat[i];i -= i & -i;}return sum;}//0-indexedint query(int l, int r) {return query(r - 1) - query(l - 1);}//[l,r)};//size, 0-indexedint n, k, m;int a[4000];int dp[4000][4000];signed main() {cin >> n >> k >> m;rep(i, n) cin >> a[i];rep(i, n)rep(j, k + 1) dp[i][j] = -inf;rep(i, n) {int sum = a[i];for (int j = i - 1; j >= i - m && j >= -1; j--) {if (j == -1) {chmax(dp[i][1], abs(sum));break;}for (int z = max(1ll, -(n - i - 1) + k); z <= k - (n - i + m - 2) / m; z++) {chmax(dp[i][z], dp[j][z - 1] + abs(sum));}//z<k-(n-i-1)//最後の場所sum += a[j];}}cout << dp[n - 1][k] << endl;}