题意
给你一个长度为(n)的数组(a)和3个数字(h,l和r)。(t)初始为0，每次可以使(t=(t+a_i) \% h)或者(t=(t+a_i-1)\%h)，如果这时(tinleft[l,r ight])就将(ans)加1。求(ans)的最大值。
解题思路
这场比赛的题感觉偏简单了。
这是一道显而易见的DP题。(dp_{i,j,k})表示枚举到(a_i)，当前(t=j)，是否-1时的(ans)的最大值，很容易就能推导出转移公式。
AC代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
#define x first
#define y second
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define endl '
'
const double PI=acos(-1.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l,int r){return l+rng()%(r-l+1);}
namespace IO{
bool REOF = 1; //为0表示文件结尾
inline char nc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && REOF && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? (REOF = 0, EOF) : *p1++;
}
template<class T>
inline bool read(T &x) {
char c = nc();bool f = 0; x = 0;
while (c<'0' || c>'9')c == '-' && (f = 1), c = nc();
while (c >= '0'&&c <= '9')x = (x << 3) + (x << 1) + (c ^ 48), c = nc();
if(f)x=-x;
return REOF;
}
template<typename T, typename... T2>
inline bool read(T &x, T2 &... rest) {
read(x);
return read(rest...);
}
inline bool need(char &c) { return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) || ((c >= 'A') && (c <= 'Z')); }
// inline bool need(char &c) { return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) || ((c >= 'A') && (c <= 'Z')) || c==' '; }
inline bool read_str(char *a) {
while ((*a = nc()) && need(*a) && REOF)++a; *a = ' ';
return REOF;
}
inline bool read_dbl(double &x){
bool f = 0; char ch = nc(); x = 0;
while(ch<'0'||ch>'9') {f|=(ch=='-');ch=nc();}
while(ch>='0'&&ch<='9'){x=x*10.0+(ch^48);ch=nc();}
if(ch == '.') {
double tmp = 1; ch = nc();
while(ch>='0'&&ch<='9'){tmp=tmp/10.0;x=x+tmp*(ch^48);ch=nc();}
}
if(f)x=-x;
return REOF;
}
template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<'='<<h<<endl; }
template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {
while(*sdbg!=',')cerr<<*sdbg++;
cerr<<'='<<h<<','<<' '; _dbg(sdbg+1, a...);
}
template<class T> ostream &operator<<(ostream& os, vector<T> V) {
os << "["; for (auto vv : V) os << vv << ","; return os << "]";
}
template<class T> ostream &operator<<(ostream& os, set<T> V) {
os << "["; for (auto vv : V) os << vv << ","; return os << "]";
}
template<class T> ostream &operator<<(ostream& os, map<T,T> V) {
os << "["; for (auto vv : V) os << vv << ","; return os << "]";
}
template<class L, class R> ostream &operator<<(ostream &os, pair<L,R> P) {
return os << "(" << P.st << "," << P.nd << ")";
}
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
}
using namespace IO;
const int maxn=2e5+5;
const int maxv=2e5+5;
const int mod=998244353; // 998244353 1e9+7
const int INF=1e9+7; // 1e9+7 0x3f3f3f3f 0x3f3f3f3f3f3f3f3f
const double eps=1e-12;
int dx[4]={0,1,0,-1};
//int dx[8]={1,0,-1,1,-1,1,0,-1};
int dy[4]={1,0,-1,0};
//int dy[8]={1,1,1,0,0,-1,-1,-1};
// #define ls (x<<1)
// #define rs (x<<1|1)
// #define mid ((l+r)>>1)
// #define lson ls,l,mid
// #define rson rs,mid+1,r
// int tot,head[maxn];
// struct Edge{
// int v,nxt;
// Edge(){}
// Edge(int _v,int _nxt):v(_v),nxt(_nxt){}
// }e[maxn<<1];
// void init(){
// tot=1;
// memset(head,0,sizeof(head));
// }
// void addedge(int u,int v){
// e[tot]=Edge(v,head[u]); head[u]=tot++;
// e[tot]=Edge(u,head[v]); head[v]=tot++;
// }
// void addarc(int u,int v){
// e[tot]=Edge(v,head[u]); head[u]=tot++;
// }
/**
* ********** Backlight **********
* 仔细读题
* 注意边界条件
* 记得注释输入流重定向
* 没有思路就试试逆向思维
* 加油，奥利给
*/
int n,h,l,r,a[maxn];
int dp[2005][2005][2];
void solve(){
read(n,h,l,r);
for(int i=1;i<=n;i++)read(a[i]);
memset(dp,-1,sizeof(dp));
dp[0][0][0]=dp[0][0][1]=0;
int sleep,delta;
for(int i=1;i<=n;i++){
for(int j=0;j<h;j++){
if(dp[i-1][j][0]!=-1){
sleep=(j+a[i])%h;
delta=(sleep>=l && sleep<=r)?1:0;
dp[i][sleep][0]=max(dp[i][sleep][0],dp[i-1][j][0]+delta);
sleep=(j+a[i]-1)%h;
delta=(sleep>=l && sleep<=r)?1:0;
dp[i][sleep][1]=max(dp[i][sleep][1],dp[i-1][j][0]+delta);
}
if(dp[i-1][j][1]!=-1){
sleep=(j+a[i])%h;
delta=(sleep>=l && sleep<=r)?1:0;
dp[i][sleep][0]=max(dp[i][sleep][0],dp[i-1][j][1]+delta);
sleep=(j+a[i]-1)%h;
delta=(sleep>=l && sleep<=r)?1:0;
dp[i][sleep][1]=max(dp[i][sleep][1],dp[i-1][j][1]+delta);
}
}
}
int ans=0;
for(int i=0;i<h;i++){
ans=max(ans,dp[n][i][0]);
ans=max(ans,dp[n][i][1]);
}
printf("%d
",ans);
}
int main()
{
// freopen("in.txt","r",stdin);
// ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
// int _T; read(_T); for(int _=1;_<=_T;_++)solve();
// while(read(n))solve();
solve();
return 0;
}