4 solutions

  • 4
    @ 2024-6-7 19:57:57
    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
        int R;
        string s;
        int p=1;
        cin>>R>>s;
        int res=0;
        reverse(s.begin(),s.end()); //反正 
        for(int i=0;i<s.size();i++)
        {
            int t; //当前位表示的值 
            if(s[i]>='A')
            {
                t=s[i]-'A'+10;
            }
            else
            {
                t=s[i]-'0';
            }
            res+=t*p; //值乘以权重 
            p=p*R; //计算下一位的权重 
        }
        cout<<res;
        return 0;
    }
    
    • 1
      @ 2025-7-22 15:52:57

      #include<bits/stdc++.h> using namespace std; int main() { int R; string s; int p=1; cin>>R>>s; int res=0; reverse(s.begin(),s.end()); //反正 for(int i=0;i<s.size();i++) { int t; //当前位表示的值 if(s[i]>='A') { t=s[i]-'A'+10; } else { t=s[i]-'0'; } res+=tp; //值乘以权重 p=pR; //计算下一位的权重 } cout<<res; return 0; }

      • 1
        @ 2025-7-22 15:34:11
        #include<bits/stdc++.h>
        using namespace std;
        
        int main()
        {
        	int su=0;
        	int r;
        	string s;
        	cin>>r;
        	cin>>s;
        	int t;
        	int jishu=1;
        	for(int i=s.size()-1;i>=0;i--)
        	{
        		if(s[i]>='A') t=s[i]-'A'+10;
        		else t=s[i]-'0';
        		su+=t*jishu;
        		jishu*=r;
        	}
        	cout<<su;
        	return 0;
        }
        
        
        
        
        • -1
          @ 2024-9-11 20:45:01
          #include<bits/stdc++.h>
          using namespace std;
          
          int main()
          {
          	int su=0;
          	int r;
          	string s;
          	cin>>r;
          	cin>>s;
          	int t;
          	int jishu=1;
          	for(int i=s.size()-1;i>=0;i--)
          	{
          		if(s[i]>='A') t=s[i]-'A'+10;
          		else t=s[i]-'0';
          		su+=t*jishu;
          		jishu*=r;
          	}
          	cout<<su;
          	return 0;
          }
          
          
          
          • 1

          Information

          ID
          105
          Time
          1000ms
          Memory
          256MiB
          Difficulty
          3
          Tags
          (None)
          # Submissions
          77
          Accepted
          26
          Uploaded By