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先標記出哪些地方有石子，然後從後往前跑dp

狀態轉移方程：dp[i] = min(dp[i],dp[i+j])    s<=j<=t;找到之前最少的需要踩的石子數，然後再加上當前是否有石子

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

const int Maxn = 100015;
const int Inf = 0x3f3f3f;
int dp[Maxn];
int vis[Maxn];
int L,s,t,n;

int main()
{
    while(~scanf("%d%d%d%d",&L,&s,&t,&n))
    {
        memset(dp,0,sizeof(dp));
        memset(vis,0,sizeof(vis));
        int index;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&index);
            vis[index] = 1;
        }

        for(int i=L;i>=0;i--)
        {
            dp[i] = Inf;
            for(int j=s;j<=t;j++)
            {
                dp[i] = min(dp[i],dp[i+j]);
            }
            dp[i] += vis[i];
        }

        printf("%d\n",dp[0]);
    }
}