CodeVs天梯之Silver

2017.12.18 By gwj1139177410

0x01排序


#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;
}

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

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;
}

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

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;
}

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;
}

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

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;
}

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

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;
}

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

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;
}

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;
}

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;
}

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;
}

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;
}