int DAC(int l,int r,int cost){
    if(r-l<=2||cost<0) return 0;
    int ans=0;
    for(int i=l+1;i<=r-2;i+=2){
        int temp=0;
        if(pre[i]-pre[l-1]==(i-l+1)>>1&&(cost-abs(a[i+1]-a[i])>=0)){
            int num1=DAC(l,i,cost-abs(a[i+1]-a[i]));
            int num2=DAC(i+1,r,cost-abs(a[i+1]-a[i]));
            temp=num1+num2+1;
            cout<<l<<" "<<r<<" at "<<i<<" with value "<<temp<<endl;
            ans=max(ans,temp);
        }
    }
    return ans;
}

一个失败的递归！因为两个子问题是互相影响的，引以为戒！！