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=500010,inf=1e9;
int n,m,x,y,z,tot;
char c;
int main()
{
    int daxie=1;
    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=0;
            }
            else 
            {
                if('a'<=c&&c<='z')putchar(c);
                else putchar(c-'A'+'a');
            }
        }
        else if(c=='.')daxie=1,putchar('.');
        else putchar(' ');
    }
    return 0;
}

　　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=500010,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=1;k=0;char c=getchar();
    while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar();
    while(c<='9'&&c>='0')k=k*10+c-'0',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]]=0;
}
int main()
{
    read(n);
    for(int i=1;i<=n;i++)read(a[i]),pre[i]=i-1,nxt[i]=i+1,q.push((poi){i,a[i]});
    for(int i=1;i<=n>>1;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]]=1;del(t.pos);
    }
    return 0;
}

　　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=20,inf=1e9;
int n,m,x,y,z,ans;
int mp[maxn][maxn],mi[maxn];
void read(int &k)
{
    int f=1;k=0;char c=getchar();
    while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar();
    while(c<='9'&&c>='0')k=k*10+c-'0',c=getchar();
    k*=f;
}
int main()
{
    read(n);read(m);
    for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)read(mp[i][j]);
    mi[0]=1;for(int i=1;i<=max(n,m);i++)mi[i]=mi[i-1]*3;
    for(int s=0;s<=mi[m];s++)
    {
        int now=0;
        for(int i=1;i<=n;i++)
        {
            int tmp1,tmp2,tmp3;tmp1=tmp2=tmp3=0;
            for(int j=1;j<=m;j++)
            if(s/mi[j-1]%3==1)
            {
                tmp1+=(mp[i][j]==2||mp[i][j]==5||mp[i][j]==8||mp[i][j]==11);
                tmp2+=(mp[i][j]==1||mp[i][j]==4||mp[i][j]==7||mp[i][j]==10);
                tmp3+=(mp[i][j]==0||mp[i][j]==3||mp[i][j]==6||mp[i][j]==9);
            }
            else if(s/mi[j-1]%3==0)
            {
                tmp1+=(mp[i][j]==3||mp[i][j]==6||mp[i][j]==9||mp[i][j]==12);
                tmp2+=(mp[i][j]==2||mp[i][j]==5||mp[i][j]==8||mp[i][j]==11);
                tmp3+=(mp[i][j]==1||mp[i][j]==4||mp[i][j]==7||mp[i][j]==10);
            }
            else if(s/mi[j-1]%3==2)
            {
                tmp1+=(mp[i][j]==1||mp[i][j]==4||mp[i][j]==7||mp[i][j]==10);
                tmp2+=(mp[i][j]==0||mp[i][j]==3||mp[i][j]==6||mp[i][j]==9);
                tmp3+=(mp[i][j]==-1||mp[i][j]==2||mp[i][j]==5||mp[i][j]==8);
            }
            now+=max(tmp1,max(tmp2,tmp3));
        }
        ans=max(ans,now);
    }
    printf("%d
",ans);
    return 0;
}