https://codeforces.ml/contest/1324/problem/E
El significado de problemas: un día h (3 <= h <= 2000) horas, n (1 <= n <= 2000) las horas de sueño, cada vez a través de la AI (1 <= ai <h) tiempo para dormir, hay dos seleccionan a través de la aI-1 ai con el tiempo para dormir o tiempo de sueño, tiempo de sueño listo para comenzar [l, r] (0 <= l <= r <h) ser buen sueño. P. Hasta unos pocos buen sueño.
Solución: dp de dos dimensiones [i] [j] representa el i-ésimo del sueño, el tiempo es de la mayor j buen sueño.
// # incluir <bits / STDC ++. h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algoritmo>
#include <iostream>
#include <string>
#include <stdio.h>
#include <cola>
#include <pila>
#include <mapa>
#include <set>
#include <string.h>
#include <vector>
typedef largo largo ll;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep (i, j, n) for (int i = j; i <= n; i ++)
#define rojo (i, n, j) for (int i = n; i> = j; i--)
#define mE (x, y) memset (x, y, sizeof (x))
// ll lcm (ll a,
// ll quickpow (ll a, ll b) {ll ans = 1, mientras que (b) {if (b & 1) ans = ans * a mod%; b >> = 1, a = a * a mod%;} return ans;}
// int euler1 (int x) {int ans = x; for (int i = 2; i * i <= x; i ++) si (x% i == 0) {Respuesta-= ans / i; mientras que (x% i == 0) x / = i;} if (x> 1) Respuesta-= American National Standard / x; ans de retorno;}
// const int N = 1E7 + 9; int VIS [n], el primer [n], phi [N]; int euler2 (int n) {ME (vis, true); int len = 1; rep (i, 2, n) {if (vis [i] ) {prime [len ++] = i, phi [i] = i-1;} for (int j = 1; j <len && prime [j] * i <= n; j ++) {vis [i * prime [j]] = 0; si (% prime i [j] == 0) {phi [i * prime [j]] = phi [i] * prime [j]; break;} else {phi [i * prime [j]] = phi [i] * phi [prime [j]];}}} len retorno}
#define INF 0x3f3f3f3f
#define acos PI (-1)
#define pii par <int, int>
#define fi primera
#define SE segundo
# definir LSON l, mediados, raíz << 1
#define rson mediados + 1, r, raíz <<
#define mp make_pair
#define cin (x) scanf ( "% LLD", y x);
using namespace std;
int const N = 1E7 + 9;
const int maxN = 2E3 + 9;
const doble esp = 1e-6;
int dp [maxN] [maxN];
int a [maxN];
anular resolver () {
int n, h, l, r;
cin >> n >> h >> l >> r;
rep (i, 1, n) {
cin >> a [i];
}
Rep (i, 0, n) {
rep (j, 0, h-1) {
dp [i] [j] = -1;
}
}
Dp [0] [0] = 0;
for (int i = 1; i <= n; i ++) {
for (int j = 0; j <h; j ++) {
si (dp [i-1] [j]!
si (k> = l && k <= r) {
dp [i] [k] = max (dp [i] [k], dp [i-1] [j] + 1);
} else {
dp [i] [k] = max (dp [i] [k], dp [i-1] [j]);
}
K = (j + a [i] - 1)% h;
si (k> = l && k <= r) {
dp [i] [k] = max (dp [i] [k], dp [i-1] [j] + 1);
} else {
dp [i] [k] = max (dp [i] [k], dp [i-1] [j]);
}
}
}
}
Int ans = -1;
for (int i = 0; i <h; i ++) {
ans = max (ans, dp [n] [i]);
}
Cout << ans << endl;
}
{
firmado main ()
// ios :: sync_with_stdio (false);
//cin.tie(0); cout.tie (0);
// int t;
// cin >> t;
// tiempo (t -) {
solve ();
//}
}
Memoria de la recursión:
#include <bits / STDC ++. h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algoritmo>
#include <iostream>
#include <string>
#include <stdio.h>
#include <cola >
#include <pila>
#include <mapa>
#include <set>
#include <string.h>
#include <vector>
typedef largo largo ll;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep (i, j, n) for (int i = j; i <= n; i ++)
#define rojo (i, n, j) for (int i = n; i> = j; i--)
#define mE (x, y) memset (x, y, sizeof (x))
// ll lcm (ll a,
ll quickpow (ll a, ll b) {ll ans = 1, mientras que (b) {if (b & 1) ans = ans * a mod%; b >> = 1, a = a * a mod%;} ans de retorno; }
// int euler1 (int x) {int ans = x; for (int i = 2; i * i <= x; i ++) si (x% i == 0) {Respuesta-= ans / i; while ( x% i == 0) x / = i;} if (x> 1) Respuesta-= American National Standard / x; ans de retorno;}
// const int N = 1E7 + 9; int VIS [n], el primer [n], phi [N]; int euler2 (int n) {ME (vis, true); int len = 1; rep (i, 2, n) {if (vis [i] ) {prime [len ++] = i, phi [i] = i-1;} for (int j = 1; j <len && prime [j] * i <= n; j ++) {vis [i * prime [j]] = 0; si (% prime i [j] == 0) {phi [i * prime [j]] = phi [i] * prime [j]; break;} else {phi [i * prime [j]] = phi [i] * phi [prime [j]];}}} len retorno}
#define SC scanf
#define INF 0x3f3f3f3f
#define acos PI (-1)
#define pii par <int, int>
#define fi primera
# definir SE segundo
#define LSON l, mediados, raíz << 1
#define rson mediados + 1, r,
using namespace std;
int const N = 1E6 + 100;
const int maxN = 2E3 + 9;
int dp [maxN] [maxN];
int a [maxN];
int n, h, l, r;
dfs int (pos int, int suma = 0) {
si (pos> n) return 0;
int & ret = dp [pos] [suma];
si (r = -1!) de retorno dp [pos] [suma];
int c = (suma + a [pos] - 1)% H;
int d = (suma + a [pos])% h;
int a = DFS (pos + 1, c) + (c> = l && c <= r);
int b = dfs (pos + 1, d) + (d> = l && d <= r);
regreso dp [pos] [suma] = max (a, b);
}
Void resolver () {
cin >> n >> h >> l >> r;
rep (i, 1, n) {
cin >> a [i];
}
rep (j, 0, h-1) {
dp [i] [j] = -1;
}
}
Tribunal << dfs (1) << endl;
}
Firmado main ()
{
// ios :: sync_with_stdio (false);
//cin.tie(0); cout.tie (0);
// int t;
// cin >> t;
// tiempo (t -) {
solve ();
//}
}