CodeVs天梯之Silver

CodeVs天梯之Silver

2017.12.18 By gwj1139177410

0x01排序

1. 明明的随机数

   
   #include<iostream>
   
   using namespace std;
   int n, a[1010], t;
   int main(){
      cin>>n;
      for(int i = 1; i <= n; i++){
          int x;  cin>>x;
          if(!a[x]){ a[x]++; t++; }
      }
      cout<<t<<"
   ";
      for(int i = 1; i <= 1000; i++)
          if(a[i])cout<<i<<" ";
      return 0;
   }

2. 排序

   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   int n, a[100010];
   int main(){
      cin>>n;
      for(int i = 0; i < n; i++)cin>>a[i];
      sort(a,a+n);
      for(int i = 0; i < n; i++)cout<<a[i]<<" ";
      return 0;
   }

0x02模拟

1. Cantor表

   
   #include<iostream>
   
   using namespace std;
   int main(){
      int n, k=1;  cin>>n;
      //1.第n个数在第k条斜线上（前k条斜线的数的个数为等差数列）
      while((1+k)*k/2 < n)k++;
      int s = n-(1+k-1)*(k-1)/2;
      //2.偶数从上往下
      if(k%2==0)cout<<s<<"/"<<k+1-s<<"
   ";
      else cout<<k+1-s<<"/"<<s<<"
   ";
      return 0;
   }

2. 蛇形矩阵

   
   #include<iostream>
   
   using namespace std;
   int a[110][110];
   int main(){
      int n;  cin>>n;
      int tot, x, y;  a[x=n][y=n] = tot = n*n;
      while(tot > 1){
          while(y-1>=1 && !a[x][y-1])a[x][--y] = --tot;
          while(x-1>=1 && !a[x-1][y])a[--x][y] = --tot;
          while(y+1<=n && !a[x][y+1])a[x][++y] = --tot;
          while(x+1<=n && !a[x+1][y])a[++x][y] = --tot;
      }
      int ans = 0;
      for(int i = 1; i <= n; i++){
          for(int j = 1; j <= n; j++){
              cout<<a[i][j]<<" ";
              if(i==j||i+j==n)ans += a[i][j];
          }
          cout<<"
   ";
      }
      cout<<ans<<"
   ";
      return 0;
   }

0x03数论入门

1. 最大公约数和最小公倍数问题

   
   #include<iostream>
   
   using namespace std;
   int gcd(int a, int b){return b==0?a:gcd(b,a%b);}
   int x, y, z, ans;
   int main(){
      cin>>x>>y; z=x*y;
      for(int i = 1; i <= z; i++)
          if(z%i==0 && gcd(i,z/i)==x)
              ans++;
      cout<<ans<<"
   ";
      return 0;
   }

2. 最大公约数

   
   #include<iostream>
   
   using namespace std;
   int gcd(int a, int b){
      return !b ? a : gcd(b,a%b);
   }
   int main(){
      int x, y;  cin>>x>>y;
      cout<<gcd(x,y)<<"
   ";
      return 0;
   }

3. 素数判定

   
   #include<iostream>
   
   using namespace std;
   int main(){
      int n;  cin>>n;
      for(int i = 2; i < n; i++)
          if(n%i==0){cout<<"\n"; return 0;}
      cout<<"\t";
      return 0;
   }

0x04进制转换

1. 十进制转m进制

   
   #include<iostream>
   
   
   #include<string>
   
   using namespace std;
   string s="0123456789ABCDEF";
   void dfs(int a, int b){
      if(a == 0)return ;
      else dfs(a/b,b);
      cout<<s[a%b];
   }
   int main(){
      int a, b;
      cin>>a>>b;
      dfs(a,b);
      return 0;
   }

2. m进制转十进制

   
   #include<iostream>
   
   
   #include<string>
   
   using namespace std;
   int main(){
      string s;  int m, t=1, ans=0;
      cin>>s>>m;
      for(int i = s.size()-1; i >= 0; i--){
          if(s[i]>='A'&&s[i]<='Z')ans += (s[i]-'A'+10)*t;
          else ans += (s[i]-'0')*t;
          t *= m;
      }
      cout<<ans<<"
   ";
      return 0;
   }

0x05递推

1. 数的计算

   
   #include<iostream>
   
   using namespace std;
   const int maxn = 1010;
   int f[maxn];
   int main(){
      int n;  cin>>n;
      for(int i = 1; i <= n; i++){
          f[i] = 1;  //左边不加也是一种
          for(int j = 0; j <= i/2; j++)f[i] += f[j];
      }
      cout<<f[n]<<"
   ";
      return 0;
   }

2. Fibonacci数列 3

   
   #include<iostream>
   
   using namespace std;
   const int maxn = 20;
   int f[maxn];
   int main(){
      int n;  cin>>n;
      f[1] = f[0] = 1;
      for(int i = 3; i <= n; i++)
          f[i%2]=f[(i-1)%2]+f[(i-2)%2];
      cout<<f[n%2];
      return 0;
   }

0x06递归

1. 二叉树最大宽度和高度

   
   #include<iostream>
   
   
   #include<algorithm>
   
   using namespace std;
   const int maxn = 110;
   int tree[maxn][2], higt, weigt[maxn], ww;//宽度是每一层的
   void dfs(int now, int dep){
      higt = max(higt,dep);
      ww = max(ww, weigt[dep]);
      if(tree[now][0]){ dfs(tree[now][0], dep+1); weigt[dep+1]++; }
      if(tree[now][1]){ dfs(tree[now][1], dep+1); weigt[dep+1]++; }
   }
   int main(){
      int n;  cin>>n;
      for(int i = 1; i <= n; i++)
          cin>>tree[i][0]>>tree[i][1];
      dfs(1, 1);
      cout<<ww+1<<" "<<higt<<"
   ";
      return 0;
   }

2. 递归第一次

   
   #include<iostream>
   
   using namespace std;
   int f(int x){
      return x>=0 ? 5 : f(x+1)+f(x+2)+1;
   }
   int main(){
      int n;  cin>>n;
      cout<<f(n)<<"
   ";
      return 0;
   }

3. 3n+1问题

   //-1什么不存在的。
   
   #include<iostream>
   
   using namespace std;
   int main(){
      int T;  cin>>T;
      while(T--){
          int n, s=0;  cin>>n;
          while(n != 1){
              if(n%2==1)n = 3*n+1;
              else n = n/2;
              s++;
          }
          cout<<s<<"
   ";
      }
      return 0;
   }

4. 二叉树的序遍历

   
   #include<iostream>
   
   using namespace std;
   const int maxn = 110;
   int tree[maxn][2];
   void dfs1(int now){
      cout<<now<<" ";
      if(tree[now][0])dfs1(tree[now][0]);
      if(tree[now][1])dfs1(tree[now][1]);
   }
   void dfs2(int now){
      if(tree[now][0])dfs2(tree[now][0]);
      cout<<now<<" ";
      if(tree[now][1])dfs2(tree[now][1]);
   }
   void dfs3(int now){
      if(tree[now][0])dfs3(tree[now][0]);
      if(tree[now][1])dfs3(tree[now][1]);
      cout<<now<<" ";
   }
   int main(){
      int n;  cin>>n;
      for(int i = 1; i <= n; i++)
          cin>>tree[i][0]>>tree[i][1];
      dfs1(1);  cout<<"
   ";
      dfs2(1);  cout<<"
   ";
      dfs3(1);  cout<<"
   ";
      return 0;
   }

5. 汉诺塔游戏

   //把大象装进冰箱一共需要几步。。。
   
   #include<iostream>
   
   using namespace std;
   int n, ans = 0, t;
   void f(int a, char b, char c){
      ans++;
      if(a == 1){
          if(t)cout<<a<<" from "<<b<<" to "<<c<<"
   ";
          return ;
      }
      f(a-1,b,198-b-c);//1.打开冰箱门
      if(t)cout<<a<<" from "<<b<<" to "<<c<<"
   ";//2.把大象装进去
      f(a-1,198-b-c,c);//关上冰箱门
   }
   int main(){
      cin>>n;
      f(n,'A','C');
      cout<<ans<<"
   ";
      t = 1;
      f(n,'A','C');
      return 0;
   }