https://codeforces.ml/contest/1324/problem/E
题意：一天有h(3 <= h <= 2000)个小时，n(1 <= n <= 2000)次睡觉，每次过ai(1 <= ai < h)时间就睡觉，有两种选择在过ai-1时间去睡或过ai时间去睡 ， 准备开始睡觉时间在[l , r]( 0 <=l <= r < h)算是好的睡眠。问最多有几个好的睡眠时间。
解法：二维dp[i][j]表示第i次睡眠，时间为j的最大好睡眠。
//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(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++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[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;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define cin(x) scanf("%lld" , &x);
using namespace std;
const int N = 1e7+9;
const int maxn = 2e3+9;
const double esp = 1e-6;
int dp[maxn][maxn];
int a[maxn];
void solve(){
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++){
if(dp[i-1][j] != -1){
int k = (j + a[i])%h;
if(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 ;
if(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;
}
signed main()
{
//ios::sync_with_stdio(false);
//cin.tie(0); cout.tie(0);
//int t ;
//cin >> t ;
//while(t--){
solve();
//}
}
记忆化递归：
#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
ll quickpow(ll a , ll b){ll ans=1;while(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++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[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;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
using namespace std;
const int N = 1e6+100;
const int maxn = 2e3+9;
int dp[maxn][maxn];
int a[maxn];
int n , h , l , r;
int dfs(int pos , int sum = 0){
if(pos > n) return 0;
int& ret = dp[pos][sum];
if(ret != -1) return dp[pos][sum];
int c = (sum + a[pos] - 1) % h ;
int d = (sum + a[pos]) % h ;
int a = dfs(pos + 1 , c) + (c >= l && c <= r) ;
int b = dfs(pos + 1 , d) + (d >= l && d <= r) ;
return dp[pos][sum] = max(a , b) ;
}
void solve(){
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 ;
}
}
cout << dfs(1) << endl;
}
signed main()
{
//ios::sync_with_stdio(false);
//cin.tie(0); cout.tie(0);
//int t ;
//cin >> t ;
//while(t--){
solve();
//}
}