```cpp
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<climits>
using namespace std;
int f[10005],far[10005],a[10005],flag[10005],p,s,t,n;
int main()
{
scanf("%d",&p);
scanf("%d%d%d",&s,&t,&n);
if(s==t) //特殊情况判断
{
int cont=0,qaq;
for(int i=1;i<=n;++i)scanf("%d",&qaq),cont+=((qaq%s)==0);
printf("%d\n",cont);return 0;
}
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
sort(a+1,a+n+1);a[0]=0;f[0]=0;
far[n+1]=min(p-a[n],100);p=0; //计算终点与最后一个点的距离
for(int i=1;i<=n;i++)far[i]=min(a[i]-a[i-1],90),p+=far[i],flag[p]=1; //缩短路径，存储缩短后的终点距离并标记石头位置
p+=far[n+1];
for(int i=1;i<=p+9;i++)
{
f[i]=INT_MAX-1;
for(int j=s;j<=t;j++)if(i>=j)f[i]=min(f[i],f[i-j]+flag[i]);
}
int minn=INT_MAX-1;
for(int i=p;i<=p+9;i++) //因为青蛙可以跳出边界且t<=10因此再终点后p-p+9中取最小值
minn=min(minn,f[i]);
printf("%d",minn);
}
by wstjy @ 2022-12-12 22:15:28