A - A Blend of Springtime

题意:给你一个字符串,字母表示花的种类.规定:花凋谢的时候可以将相邻两个地方给染色,问是否存在一个地方有所有的颜色.

思路:只要有三个连续的位置由三个不同的字母填充就可以.

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;


int main()
{
    string s;
    cin>>s;
    int l = s.length();
    if(l<3)
    {
        cout<<"No\n";
        return 0;
    }
    int flag = 0;
    //int i = 0;
    for(int i = 0 ;i <= l-3;i++)
    {
        if(s[i]!=s[i+1]&&s[i+1]!=s[i+2]&&s[i]!=s[i+2]&&s[i]!='.'&&s[i+1]!='.'&&s[i+2]!='.')
        {
            flag = 1;
            break;
        }
    }
   // cout<< 'A'+'C'+'B';
    if(flag)
        cout<<"Yes\n";
    else cout<<"No\n";
}

B - A Tide of Riverscape

题意:给你两个数字n和p,分别表示字符串的长度和周期,给你一个由0,1和逗号组成的字符串,让你判断是否任意间隔p个的字符的两个字符不相同,如果都不相同输出no,反之输出yes,并且.可以转化为0或者一.

思路:用滑动窗口的方法 去遍历字符串就行

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

int main()
{
    int n,p;
    string s;
    cin>>n>>p;
    cin>>s;
    LL num;
        int flag = 0;
        for( int  i = 0; i < s.length(); i ++ )
        {
            if( i+p < s.length() )
            {
                if( s[i] != s[i+p]||s[i] == s[i+p] && s[i] == '.')
                {
                    if(s[i] == s[i+p])
                    {
                        s[i]='0', s[i+p] = '1';
                    }
                    else
                    {
                        if(s[i] == '0')
                        {
                            s[i+p] = '1';
                        }
                        else if(s[i] == '1')
                        {
                            s[i+p] = '0';
                        }
                        else
                        {
                            if(s[i+p] == '0')
                            {
                                s[i] = '1';
                            }
                            else
                            {
                                s[i] = '0';
                            }
                        }
                    }flag = 1;
                    break;
                }
            }
        }
        if(flag)
        {
            for(int i = 0; i < s.length();i++)
            {
                if(s[i] == '.')
                {
                    s[i] = '0';
                }
            }
            cout<<s<<endl;
        }
        else
        {
            cout<<"No";
        }

}

D - Infinity Gauntlet

题意:六种颜色分别对应六种元素,给你n个颜色,问你还剩下什么元素;

思路:用map对应,然后用一个标记数组储存是否被标记过,最后输出没被标记的元素就行

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
//map<string,string> a = {{"purple","Power"},{"green","Time"},{"blue","Space"},{"orange","Soul"},{"red","Reality"},{"yellow","Mind"}};
map<string,int>a = {{"purple",0},{"green",1},{"blue",2},{"orange",3},{"red",4},{"yellow",5}};
string ccc[6] = {"Power","Time","Space","Soul","Reality","Mind"};
int main()
{
    int n;
    cin>>n;
    int res = 6-n;
    int s[6] = {0};
    for(int i = 0;i < n;i++){
        string t;
        cin>>t;
        s[a[t]] = 1;
    }

    cout<<res<<endl;
    for(int i = 0;i <=5;i++)
    {
        if(s[i]==0)
            cout<<ccc[i]<<endl;
    }

}

E - High School: Become Human

题意:让你比较x^y和y^x的大小

思路:比较对应的对数大小就行,特别判断等号:如果两个数相等或者一个等于2,另一个等于4那么两个相等.

#include<bits/stdc++.h>
using namespace std;
int main()
{
   long long a,b;
   cin>>a>>b;
   if(a==b||a==2&&b==4||a==4&&b==2)
    cout<<"="<<endl;
   else
   {
        double x = b*log(a),y = a*log(b);
       if(x>y)
        cout<<">"<<endl;
       else if(x<y)
        cout<<"<"<<endl;
   }

}