第一步：写标准的程序

#include <iostream>
using namespace std;
int main(void){
    long long n,k,d;
    register long long *a,*f;
    cin>>n>>k>>d;
    a=new long long[n+1];
    f=new long long[n+1];
    for(register unsigned int i=1;i<=n;i++){
        cin>>a[i];
        f[i]=0x6fffffffffffffff;
    }
    f[1]=a[1];
    for(register unsigned int i=2;i<=n;i++)
        for(register unsigned int j=1;j<i;j++)
            f[i]=min(f[i],f[j]+k+(i-j-1)*d+a[i]);
    cout<<f[n];
    delete[] a;
    delete[] f;
    return 0;
}

第二步：拆解状态转移方程

#include <iostream>
using namespace std;
int main(void){
    long long n,k,d,val;
    register long long *a,*f;
    cin>>n>>k>>d;
    a=new long long[n+1];
    f=new long long[n+1];
    for(register unsigned int i=1;i<=n;i++){
        cin>>a[i];
        f[i]=0x6fffffffffffffff;
    }
    f[1]=a[1];
    for(register unsigned int i=2;i<=n;i++){
        val=0x6fffffffffffffff;
        for(register unsigned int j=1;j<i;j++)
            val=min(val,f[j]+j*d);
        f[i]=val+k+(i-1)*d+a[i];
    }
    cout<<f[n];
    delete[] a;
    delete[] f;
    return 0;
}

第三步：找到规律

val1=f[1]+d

val2=min(f[1]+d,f[2]+2*d)

val3=min(f[1]+d,f[2]+2d,f[3]+3d)

所以val=min(val,f[j]+j*d)

第四步：完善代码

#include <iostream>
using namespace std;
int main(void){
    long long n,k,d,val;
    register long long *a,*f;
    cin>>n>>k>>d;
    a=new long long[n+1];
    f=new long long[n+1];
    for(register unsigned int i=1;i<=n;i++){
        cin>>a[i];
        f[i]=0x6fffffffffffffff;
    }
    f[1]=a[1];
    val=f[1]-d;
    for(register unsigned int i=2;i<=n;i++){
        f[i]=val+(k+(i-1)*d+a[i]);
        val=min(val,f[i]-i*d);
    }
    cout<<f[n];
    delete[] a;
    delete[] f;
    return 0;
}

设f[i]表示解决完第i个学生的问题后，花费的最小精力，则

$f[i]=min(f[k]+k+(i-j-1)*d+a[i]);$

按照i和j变量进行分离，

$f[i]=min(f[j]-j*d)+(k+(i-1)*d+a[i]);$

和$i$相关变量视为常数，和$j$相关变量需要维护最小值

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1000005;
ll a[N],n,m,k,s,t,d,f[N];
int main(){
	cin>>n>>k>>d;
	for (int i=1;i<=n;++i) cin>>a[i];
	f[1]=a[1];
	ll mi=f[1]-1*d;
	for (int i=2;i<=n;++i){
		f[i]=mi+k+(i-1)*d+a[i];
		mi=min(mi,f[i]-i*d);
	}
	cout<<f[n];
}