蓝桥31天|今天4道题Day24|C++

1.跑步锻炼

#include 
using namespace std;
int month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int week=4;
bool isdate(int date){
   int yyyy=date/10000;
   int mm=date/100%100;
   int dd=date%100;
   if(mm<1||mm>12)return false;
   if(yyyy%400==0||(yyyy%100!=0&&yyyy%4==0))month[2]=29;
   else month[2]=28;

   if(dd<=0||dd>month[mm])return false;
   return true;
}
bool isfirst(int date){
   int dd=date%100;
   if(dd==1)return true;
   return false;
}
bool ismon(){
   week=(week+1)%7;
   //printf("week=%d\n",week);
   if(week==0)return true;
   return false;
}
int main()
{
  long long ans=0;
  for(int i=20000101;i<=20201001;i++){
     if(isdate(i)){
        ans++;
        if(ismon()||isfirst(i)){
          ans++;
          //printf("%d\n",i);
        }

     }
  }
 
  printf("%lld",ans);
  return 0;
}

2.直线

题目地址

#include 
#include 
#include 
using namespace std;
typedef pair<double,double>PDD;
#define x first
#define y second
vector<PDD>point;
set<PDD>line;
int main(){
  int n=20,m=21;
  int sum=n+m;
  for(int i=0;i<n;i++){
    for(int j=0;j<m;j++){
      point.push_back({(double)i,(double)j});
      for(int k=0;k<i*m+j;k++){
         PDD t=point[k];
         if(t.x==i||t.y==j)continue;

         double a=(t.y-j)*1.0/(t.x-i);
         // b=(x1y2-x2y1)/(x2-x1)   截距计算，最小二乘法
         double b=(t.y*i-t.x*j)*1.0/(t.x-i);
         //printf("a==%lf b==%lf\n",a,b);
         int len=line.size();
         line.insert({a,b});
         if(len!=line.size())sum++;


      }
    }
  }
   printf("%d\n",sum);




  return 0;
}

3.循环小数

题目链接

#include
#define ll long long
using namespace std;
ll gcd(ll a,ll b)
{
    return b?gcd(b,a%b):a;
}


int main()
{
    ll p,q,num;
    cin>>p>>q>>num;
    ll X,Y;//x分子，y分母

    ll t=pow(10,q-p+1); //1e5

    ll loop=num%t;//循环数值     42857
    ll l=num/t;//未循环            1

    X=loop+(t-1)*l;         //42857+(1e5-1)*1
    Y=pow(10,p-1)*(t-1);     //10*(1e5-1)

    ll g=gcd(X,Y);
    X/=g;
    Y/=g;
    cout<<X<<" "<<Y<<endl;


    return 0;
}

4.卡牌换位

#include 
#include 
#include 
using namespace std;
string st,rst;
int a,b;
int dx[4]= {1,0,-1,0},dy[4]= {0,1,0,-1};
struct node{
    string str;//当前字符
    int p;//空字符位置
    int step;//已经走的步数
    int pre;//空字符上次位置
    node(){
    }
    node(string str,int p,int step,int pre){
        this->str=str;
        this->p=p;
        this->step=step;
        this->pre=pre;
    }
};
int bfs(string st,int p,int step)
{
    queue<node>q;

    q.push(node(st,p,0,-1));
    while(!q.empty())
    {
        node t=q.front();
        string nst=t.str;
        int step=t.step,p=t.p,pre=t.pre;
        q.pop();
        if(nst[a]=='B'&&nst[b]=='A')return step;
        int x=p/3,y=p%3;
        for(int i=0; i<4; i++)
        {
            int sx=x+dx[i],sy=y+dy[i];
            if(sx*3+sy!=pre&&sx>=0&&sx<2&&sy>=0&&sy<3)
            {
                swap(nst[sx*3+sy],nst[p]);
                q.push({nst,sx*3+sy,step+1,p});
                swap(nst[sx*3+sy],nst[p]);

            }
        }

    }
    return -1;
}
int main()
{

    getline(cin,st);
    getline(cin,rst);
    st+=rst;

    int p;
    for(int i=0; i<6; i++)
    {
        if(st[i]=='A')
            a=i;
        if(st[i]=='B')
            b=i;
        if(st[i]==' ')
            p=i;
    }

    //字符串,为空的位置,交换次数,上一个交换的位置
    cout<<bfs(st,p,0);
    return 0;
}