解

我写的：
#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);
}
DEEPSEEK写的：

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;

int main() {
int L, S, T, M;
cin >> L >> S >> T >> M;
vector<int> stones(M);
for (int i = 0; i < M; ++i) cin >> stones[i];
sort(stones.begin(), stones.end());

// 路径压缩
if (S == T) {
    int cnt = 0;
    for (int pos : stones) cnt += (pos % S == 0);
    cout << cnt << endl;
    return 0;
}

vector<int> compressed;
int prev = 0;
for (int pos : stones) {
    int diff = pos - prev;
    if (diff > S * T) diff = S * T;
    compressed.push_back(prev + diff);
    prev = compressed.back();
}

// DP初始化
vector<int> dp(prev + T + 2, INT_MAX);
dp = 0;
for (int i = 1; i <= prev + T; ++i) {
    for (int j = S; j <= T; ++j) {
        if (i - j >= 0 && dp[i - j] != INT_MAX) {
            bool isStone = binary_search(stones.begin(), stones.end(), i);
            dp[i] = min(dp[i], dp[i - j] + (isStone ? 1 : 0));
        }
    }
}

// 检查终点附近的最小值
int ans = INT_MAX;
for (int i = prev; i <= prev + T; ++i) {
    ans = min(ans, dp[i]);
}
cout << ans << endl;
return 0;

}
哪个对大家自己想