5 条题解

  • 1
    @ 2024-1-7 19:56:41
    b = [1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,4,1,2,3,1,2,3,4]
    print(sum([b[ord(i) - ord("a") if i != " " else 0] for i in input()]))
    
    • 0
      @ 2024-1-7 14:35:19
      #include<bits/stdc++.h>
      using namespace std;
      #define ll long long
      #define endl '\n'
      void solve();
      
      int main(){
      	ios::sync_with_stdio(0);
      	cin.tie(0);
      	cout.tie(0);
      	int t=1;
      //	cin>>t;
      	while(t--){
      		solve();
      	}
      	return 0;
      }
      
      void solve(){
      	string s;
      	getline(cin,s);
      	map<char,int>a;
      	for(int i='a';i<='o';i++){
      		a[i]=(i-'a')%3+1;
      	}
      	a['p']=1;
      	a['q']=2;
      	a['r']=3;
      	a['s']=4;
      	a['t']=1;
      	a['u']=2;
      	a['v']=3;
      	a['w']=1;
      	a['x']=2;
      	a['y']=3;
      	a['z']=4;
      	a[' ']=1;
      	ll ans=0;
      	for(auto c:s){
      		ans+=a[c];
      	}
      	cout<<ans<<endl;
      }
      
      • 0
        @ 2024-1-7 13:29:03
        import java.util.Scanner;
        
        public class Main {
            public static void main(String[] args) {
                Scanner sc = new Scanner(System.in);
                String str = sc.nextLine();
                int ans = 0;
                for (int i = 0; i < str.length(); i++) {
                    ans += solve(str.charAt(i));
                }
                System.out.println(ans);
            }
            public static int solve(char key){
                switch (key){
                    case 'a' :
                    case 'g' :
                    case 'j':
                    case 'd' :
                    case 'm':
                    case 'p':
                    case 't':
                    case 'w':
                    case ' ':
                        return 1;
                    case 'b' :
                    case 'e' :
                    case 'h' :
                    case 'k':
                    case 'n':
                    case 'q':
                    case 'u':
                    case 'x':
                        return 2;
                    case 'c':
                    case 'i':
                    case 'f' :
                    case 'l':
                    case 'o':
                    case 'r':
                    case 'v':
                    case 'y':
                        return 3;
                    case 's':
                    case 'z':
                        return 4;
                    default:
                        return 0;
                }
            }
        }
        
        
        • 0
          @ 2024-1-7 13:25:49
          import java.util.*;
          
          public class Main {
              public static void main(String[] args) {
                  Scanner sc = new Scanner(System.in);
                 int[] arr = {1,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,4,1,2,3,1,2,3,4};
                  char[] c = sc.nextLine().toCharArray();
                  int ans = 0;
                  for (char temp : c) {
                      //如果当前字符为空格,则ans加1
                      if(temp==' ')ans++;
                      //否则,ans加上arr中对应字符的值
                      else{
                          ans+=arr[temp-96];
                      }
                  }
                  System.out.println(ans);
              }
          }
          
          • 0
            @ 2024-1-7 13:18:46
            #include <vector>
            #include <set>
            #include <string>
            using namespace std;
            #define int long long
            const int mod=1e+7;
            int a[26]={1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,4,1,2,3,1,2,3,4};
            signed main() {
                iostream::sync_with_stdio(false);
                cin.tie(0);
                cout.tie(0);
                string str;
                getline(cin,str);
                int num=0;
                for(int i=0;i<str.size();i++){
                    if(str[i]==' '){//注意空格
                        num++;
                    }else{
                        num+=a[str[i]-'a'];
                    }
                }
                cout <<num<< endl;
                return 0;//AC
             }
            
            • 1

            信息

            ID
            589
            时间
            1000ms
            内存
            256MiB
            难度
            3
            标签
            递交数
            142
            已通过
            75
            上传者