枚举大法

一维密码锁（可能没有解）

第一个按钮按或者不按

#include<iostream>
#include<cstring>
#include<algorithm>
#include<bitset>
using namespace std;

int main()
{
    string line;
    bitset<32> lock;
    int mintimes=99999999;//无穷大 
    
    cin>>line;
    bitset<32> sourcelock(line);
    cin>>line;
    bitset<32> targetlock(line);
    
    int n=line.size();
    for(int p=0;p<2;++p)
    {
        lock=sourcelock;
        int times=0;
        int nextbutton=p;
        
        for(int i=0;i<n;++i)
        {
            if(nextbutton==1)
            {
                ++times;
                if(i>0)
                    lock.flip(i-1);
                lock.flip(i);
                if(i<n-1)
                lock.flip(i+1);
            }
            
            if(lock[i]!=targetlock[i])
                nextbutton=1;
            else
                nextbutton=0;
            if(lock==targetlock)        
            {
                mintimes=min(mintimes,times);
            }
        }
    }
        
    if(mintimes==99999999)
    {
        cout<<"impossible\n";
    }
    
    else
    {
        cout<<mintimes<<endl;
    }
}

 

二维熄灯问题

枚举第一行所有情况

#include<iostream>
#include<bitset>
#include<cstring>
#include<memory>
using namespace std;

bitset<6> source[5],lights[5],res[5],line;
int main()
{
    int T;
    cin>>T;
    for(int t=1;t<=T;++t)
    {
        for(int i=0;i<5;i++)
        {
            for(int j=0;j<6;j++)
            {
                int x;
                cin>>x;
                source[i].set(j,x);
            }
        }
    
        for(int n=0;n<64;++n)
        {
        for(int i=0;i<5;i++)    lights[i]=source[i];
        line=n;
        for(int i=0;i<5;++i)
        {
            res[i]=line;
            
            for(int j=0;j<6;++j)
            {
                if(line.test(j))
                {
                    if(j>0)        lights[i].flip(j-1);
                    lights[i].flip(j);
                    if(j<5)        lights[i].flip(j+1);
                }
            }
            
            if(i<4)        lights[i+1]^=line;
            line=lights[i];
        }
            if(lights[4].none())
            {
                cout<<"PUZZLE #"<<t<<"\n";
                for(int i=0;i<5;++i)
                {
                    for(int j=0;j<6;++j)
                    {
                        cout<<res[i][j]<<" ";
                    }
                    cout<<endl;
                }
                
            }
        }
        
    }
    
}

 

拨钟问题

九重循环枚举所有操作

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
    int a[11];
    int change[11];
    int best[11];
    int minCount = 999999;
     for(int i = 1;i<=9;++i)
     {
         cin >>a[i];
     }
     memset(change,0,sizeof(change));
     int sum = 0;

         for(change[1] = 0;change[1]<4;change[1]++)
         for(change[2] = 0;change[2]<4;change[2]++)
         for(change[3] = 0;change[3]<4;change[3]++)
         for(change[4] = 0;change[4]<4;change[4]++)
         for(change[5] = 0;change[5]<4;change[5]++)
         for(change[6] = 0;change[6]<4;change[6]++)
         for(change[7] = 0;change[7]<4;change[7]++)
         for(change[8] = 0;change[8]<4;change[8]++)
         for(change[9] = 0;change[9]<4;change[9]++)
         {
             sum = 0;
             sum+= (a[1]+change[1]+change[2]+change[4])%4;//A只有1 2 4操作能改变A ，%4拨4次相当于没有拨 
             sum+= (a[2]+change[1]+change[2]+change[3]+change[5])%4;//B
             sum+= (a[3]+change[2]+change[3]+change[6])%4;//C
             sum+= (a[4]+change[1]+change[4]+change[5]+change[7])%4;//D
             sum+= (a[5]+change[1]+change[3]+change[5]+change[7]+change[9])%4;//E
             sum+= (a[6]+change[3]+change[5]+change[6]+change[9])%4;//F
             sum+= (a[7]+change[4]+change[7]+change[8])%4;//G
             sum+= (a[8]+change[5]+change[7]+change[8]+change[9])%4;//H
             sum+= (a[9]+change[6]+change[8]+change[9])%4;//I
             if(sum == 0)//能恢复 
             {
                 int count = 0;
                 for(int j=1;j<=9;j++)
                 {
                      count += change[j];
                 }
                  if(count < minCount)
                  {
                      minCount = count;
                      memcpy(best,change,sizeof(change));
                }
                for(int j=1;j<=9;j++)
                {
                      while(best[j]--)
                      {
                          cout << j<<" ";
                      }
                }
             }
         }
    return 0;
 }