100114D

这道题用暴力水过了，蒟蒻是这么想的：枚举两个端点，找最小值，因为shift只能用一次，但是这样10^9*2.5要t，所以减掉只有一个黑点的情况，然后复杂度变为10^9*0.6

#include<iostream>
#include<vector>
#include<cstdio>
using namespace std;
int n,ll=-(<<),rr=-(<<),ans;
vector<int>left1;
vector<int>right1;
char c;
char s[];
int sum[];
int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    cin>>n;
    int pos=;
    cin.ignore();
    for(int i=;i<=n;i++)
    {
        c=getchar();
        s[i]=c;
        sum[i]=sum[i-]+(c=='*');
    }
    s[]='.';
    s[++n]='.';
    for(int i=;i<=n;i++)
    {
        int l,r;
        if(s[i]=='*'&&s[i-]=='.'){l=i;left1.push_back(i);}
        if(s[i]=='*'&&s[i+]=='.')
        {
            r=i;
            if(r-l+<)
            {
                left1.pop_back();
                continue;
            }
            right1.push_back(i);

        }
    }
    int l=,r=;
    int MAX=-(<<);
    for(int i=;i<left1.size();i++)
    {
        for(int j=i;j<right1.size();j++)
        {
            int a=right1[j]-left1[i]+;
//            cout<<"left:"<<left1[i]<<endl;
//            cout<<"right:"<<right1[j]<<endl;
            int b=sum[right1[j]]-sum[left1[i]-];//操作数a-b+2和b比较
//            cout<<"MAX="<<MAX<<" "<<a<<" "<<b<<endl;
            if(a-b+<b&&MAX<*b-a-)
            {
                MAX=*b-a-;
                ll=left1[i];rr=right1[j];
            }
        }
    }
//    cout<<"ll="<<ll<<" "<<"rr="<<rr<<endl;
    if(MAX!=-(<<))
    {
        ans+=(+rr-ll-*(sum[rr]-sum[ll-]));
    }
    ans+=sum[n-];
//    cout<<sum[n]<<endl;
    cout<<ans<<endl;
    if(MAX!=-(<<))
    {
        cout<<ll<<endl;
        cout<<"Shift+"<<rr<<endl;
        for(int i=ll;i<=rr;i++)
            if(s[i]=='.')
            {
                cout<<"Ctrl+"<<i<<endl;
            }
    }
    if(ll==-(<<)&&rr==-(<<))
    {
        ll=;rr=;
    }
    for(int i=;i<ll;i++)
        if(s[i]=='*')cout<<"Ctrl+"<<i<<endl;
    for(int i=rr+;i<=n;i++)
        if(s[i]=='*')cout<<"Ctrl+"<<i<<endl;
    fclose(stdin);
    fclose(stdout);
    return ;
}