Luo Gu P5614 solution to a problem

Make complaints:~~Data seems a little water~~Direct enumeration to 200 can get 80 points.

Another: I was too weak, and when the game only 90 points, # 7 dead card is not in the past, and finally found no judge \ (z_1 \) and \ (z_2 \) range ......

Subtask 1:

Method:

Direct output 4, finished.

Subtask 2：

Method:

Direct violence enumeration \ (X \) , \ (Y \) , \ (Z \) , determines whether or not to look at the relationship satisfies:
\ [\ FORALL I \ in \ left [. 1, n-\ right], ST \ left | a_i-x \ right | \
bigoplus \ left | b_i-y \ right | \ bigoplus \ left | c_i -z \ right | = 9 \] time complexity: \ (O (nM ^. 3) \) ( \ ( the n-\ Leq 5 \) , can be ignored)

By \ (60 \% \) data.

Code：

#include<bits/stdc++.h>
#define int long long 
#define Maxn 10
using namespace std;
inline void read(int &x)
{
    int f=1;x=0;char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
    x*=f;
}
int n,M;
int ans=0;
int a[Maxn],b[Maxn],c[Maxn];
void solve1()
{
    int flag=1;
    for(int x=1;x<=M;x++)
    {
        for(int y=1;y<=M;y++)
        {
            for(int z=1;z<=M;z++)
            {
                for(int i=1;i<=n;i++)
                {
                    if((((abs(a[i]-x))^(abs(b[i]-y)))^(abs(c[i]-z)))!=9)
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag==1)
                {
                    ans++;
                }else
                {
                    flag=1;
                }
            } 
        }
    }
    printf("%lld\n",ans);
    return ;    
} 
signed main()
{
    read(n),read(M);
    for(int i=1;i<=n;i++)
    {
        read(a[i]);
        read(b[i]);
        read(c[i]);
    }
    ans=0;
    if(M<=200)
    {
        solve1();
        return 0;
    }
}

Subtask 3:

Method :

Consider optimizing enumeration. If yes what properties:
\ [A \ B bigoplus \ bigoplus C = D \]
it will be apparent to give:
\ [\ the begin Equation} {\ the aligned the begin {A} & \ bigoplus B \ \\ & bigoplus D = A \ bigoplus B \ bigoplus \ left (A \ bigoplus B \ bigoplus C \ right) \\ & = \ left [\ left (A \ bigoplus B \ right) \ bigoplus \ left (A \ bigoplus B \ right) \ right] \ bigoplus C \\ & \ because x \
bigoplus x = 0,0 \ bigoplus x = x \\ & \ therefore A \ bigoplus B \ bigoplus D = C \ end {aligned} \ end {equation} \] requires only pieces For \ (X \) , \ (Y \) , then:
\ [\ the begin {Equation} \ the begin {the aligned} & \ left | c_1 and-Z \ right | = \ left | A_1-X \ right | \ bigoplus \ left | b_1-y \ right |
\ bigoplus 9 \ end {aligned} \ end {equation} \] we set
\ [C = \ left | a_1 -x \ right | \ bigoplus \ left | b_1-y \ right | \ bigoplus 9 \]
is:
\ [Z_1 = c_1 and-C, C + c_1 and Z_2 = \]
Finally, only we need to determine what \ (z_1 \) and \ (z_2 \) satisfies the following properties can be:
\ [\ FORALL i \ in \ left [2, the n-\ right], ST \ left | a_i-the X-\ right | \ bigoplus \ left | b_i-y \ right | \ bigoplus \ left | c_i -z_j \ right | = 9 \ text { and} z_j \ in [1, M ] \\ (j \ in \ {1,2 \} ) \]
time complexity: \ (O (nM 2 ^) \) ( \ (n-\ Leq. 5 \) , can be ignored)

Code:

#include<bits/stdc++.h>
#define int long long 
#define Maxn 10
using namespace std;
inline void read(int &x)
{
    int f=1;x=0;char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
    x*=f;
}
int n,M;
int ans=0;
int a[Maxn],b[Maxn],c[Maxn];
map<int,int>mp;
void solve1()
{
    int flag=1;
    for(int x=1;x<=M;x++)
    {
        for(int y=1;y<=M;y++)
        {
            for(int z=1;z<=M;z++)
            {
                for(int i=1;i<=n;i++)
                {
                    if((((abs(a[i]-x))^(abs(b[i]-y)))^(abs(c[i]-z)))!=9)
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag==1)
                {
                    ans++;
                }else
                {
                    flag=1;
                }
            } 
        }
    }
    printf("%lld\n",ans);
    return ;    
} 
void solve2()
{
    int flag1=1,flag2=1;
    for(int x=1;x<=M;x++)
    {
        for(int y=1;y<=M;y++)
        {
            mp.clear();
            int tmp=(((abs(a[1]-x))^(abs(b[1]-y)))^9);
            int zkkk=c[1]-tmp;
            int zwww=c[1]+tmp;
            if(zkkk>=1&&zkkk<=M)
            {
                for(int i=2;i<=n;i++)
                {
                    if((((abs(a[i]-x))^(abs(b[i]-y)))^(abs(c[i]-zkkk)))!=9)
                    {
                        flag1=0;
                        break;
                    }
                }
                if(flag1==1&&mp.find(zkkk)==mp.end())
                {
                    ans++;
                    mp[zkkk]=1;
                }else
                {
                    flag1=1;
                }
            }else
            {
                flag1=1;
            }
            if(zwww>=1&&zwww<=M)
            {
                for(int i=2;i<=n;i++)
                {
                    if((((abs(a[i]-x))^(abs(b[i]-y)))^(abs(c[i]-zwww)))!=9)
                    {
                        flag2=0;
                        break;
                    }
                }
                if(flag2==1&&mp.find(zwww)==mp.end())
                {
                    ans++;
                }else
                {
                    flag2=1;
                }
            }else
            {
                flag2=1;
            }
        }   
    }
    printf("%lld\n",ans);
    return ;
}
signed main()
{
    read(n),read(M);
    for(int i=1;i<=n;i++)
    {
        read(a[i]);
        read(b[i]);
        read(c[i]);
    }
    ans=0;
    if(M<=200)
    {
        solve1();
        return 0;
    }
    else 
    {
        solve2();
        return 0;
    }
}