哎，这题我能说想到增加起始和终结点的嘛，一直在想dp[i]和dp[i-1]的关系，然后c就很难处理...

dp[i] = dp[j]+f(i,j)

可以说很水嘛，要哭了

//dp[i] = dp[j]+t(i,j) t(i,j)为i到j的时间//增加其实和终结点(技巧)#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               const int inf = (1<<31)-1;const int MAXN = 102;using namespace std;int a[MAXN];double dp[MAXN];bool cmp(double a,double b){ if(a-b>0)return true; else return false;}int main(){ int l,n,c,t; int vr,vt1,vt2; while(~scanf("%d",&l)){ scanf("%d%d%d",&n,&c,&t); scanf("%d%d%d",&vr,&vt1,&vt2); for(int i=1;i<=n;i++) scanf("%d",&a[i]); a[0] = 0; a[n+1] = l; double ts; for(int i=1;i<=n+1;i++){ dp[i] = inf; for(int j=i-1;j>=0;j--){ if(c>a[i]-a[j])ts = (a[i]-a[j])*1./vt1; else ts = c*1./vt1+(a[i]-a[j]-c)*1./vt2; if(j!=0)ts += t; dp[i] = min(dp[i],ts+dp[j]); } } if(cmp(dp[n+1],l*1./vr)) cout<<"Good job,rabbit!"<