浴谷八连测R4题解

　　一开始出了点bug能看见排行榜，于是我看见我半个小时就A掉了前两题，信心场QAQ

　　T1字符串题就不说了qwq

#include<iostream>

#include<cstring>

#include<cstdlib>

#include<cstdio>

#include<algorithm>

#define ll long long

using namespace std;

const int maxn=,inf=1e9;

int n,m,x,y,z,tot;

char c;

int main()

{

    int daxie=;

    while((c=getchar())!=EOF)

    {

        if(('a'<=c&&c<='z')||('A'<=c&&c<='Z'))

        {

            if(daxie)

            {

                if('a'<=c&&c<='z')putchar(c-'a'+'A');

                else putchar(c);

                daxie=;

            }

            else

            {

                if('a'<=c&&c<='z')putchar(c);

                else putchar(c-'A'+'a');

            }

        }

        else if(c=='.')daxie=,putchar('.');

        else putchar(' ');

    }

    return ;

}

　　T2直接贪心不断找出最大的数和他的下一个数，用双向链表维护一下前面和后面的数就好了

#include<iostream>

#include<cstring>

#include<cstdlib>

#include<cstdio>

#include<algorithm>

#include<queue>

#define ll long long

using namespace std;

const int maxn=,inf=1e9;

struct poi{int pos,w;};

priority_queue<poi>q;

bool operator<(poi a,poi b){return a.w<b.w;}

int n,m,x,y,z,tot;

int nxt[maxn],pre[maxn],a[maxn];

bool v[maxn];

void read(int &k)

{

    int f=;k=;char c=getchar();

    while(c<''||c>'')c=='-'&&(f=-),c=getchar();

    while(c<=''&&c>='')k=k*+c-'',c=getchar();

    k*=f;

}

void del(int x)

{

    nxt[pre[x]]=nxt[nxt[x]];

    pre[nxt[nxt[x]]]=pre[x];

    pre[x]=nxt[x]=pre[nxt[x]]=nxt[nxt[x]]=;

}

int main()

{

    read(n);

    for(int i=;i<=n;i++)read(a[i]),pre[i]=i-,nxt[i]=i+,q.push((poi){i,a[i]});

    for(int i=;i<=n>>;i++)

    {

        poi t=q.top();q.pop();

        while(v[t.pos]||nxt[t.pos]>n)t=q.top(),q.pop();

        printf("%d %d ",t.w,a[nxt[t.pos]]);

        v[nxt[t.pos]]=;del(t.pos);

    }

    return ;

}

　　T3可以发现每行每列最多加两次，因为每隔2个数就有一个稳数...于是3^n枚举列的状态，算出每一行不加，加一个，加两个哪个优，每一行互不影响，于是每一行贪心地选一个最大的就好了

#include<iostream>

#include<cstring>

#include<cstdlib>

#include<cstdio>

#include<algorithm>

#define ll long long

using namespace std;

const int maxn=,inf=1e9;

int n,m,x,y,z,ans;

int mp[maxn][maxn],mi[maxn];

void read(int &k)

{

    int f=;k=;char c=getchar();

    while(c<''||c>'')c=='-'&&(f=-),c=getchar();

    while(c<=''&&c>='')k=k*+c-'',c=getchar();

    k*=f;

}

int main()

{

    read(n);read(m);

    for(int i=;i<=n;i++)for(int j=;j<=m;j++)read(mp[i][j]);

    mi[]=;for(int i=;i<=max(n,m);i++)mi[i]=mi[i-]*;

    for(int s=;s<=mi[m];s++)

    {

        int now=;

        for(int i=;i<=n;i++)

        {

            int tmp1,tmp2,tmp3;tmp1=tmp2=tmp3=;

            for(int j=;j<=m;j++)

            if(s/mi[j-]%==)

            {

                tmp1+=(mp[i][j]==||mp[i][j]==||mp[i][j]==||mp[i][j]==);

                tmp2+=(mp[i][j]==||mp[i][j]==||mp[i][j]==||mp[i][j]==);

                tmp3+=(mp[i][j]==||mp[i][j]==||mp[i][j]==||mp[i][j]==);

            }

            else if(s/mi[j-]%==)

            {

                tmp1+=(mp[i][j]==||mp[i][j]==||mp[i][j]==||mp[i][j]==);

                tmp2+=(mp[i][j]==||mp[i][j]==||mp[i][j]==||mp[i][j]==);

                tmp3+=(mp[i][j]==||mp[i][j]==||mp[i][j]==||mp[i][j]==);

            }

            else if(s/mi[j-]%==)

            {

                tmp1+=(mp[i][j]==||mp[i][j]==||mp[i][j]==||mp[i][j]==);

                tmp2+=(mp[i][j]==||mp[i][j]==||mp[i][j]==||mp[i][j]==);

                tmp3+=(mp[i][j]==-||mp[i][j]==||mp[i][j]==||mp[i][j]==);

            }

            now+=max(tmp1,max(tmp2,tmp3));

        }

        ans=max(ans,now);

    }

    printf("%d\n",ans);

    return ;

}