| Run ID | Author | Problem | Lang | Verdict | Score | Time | Memory | Code Length | Submit Time |
|---|---|---|---|---|---|---|---|---|---|
| 325590 | 黄浙峰老师 | 【C6-3】进制转换 | C++ | Accepted | 100 | 1 MS | 260 KB | 517 | 2025-06-07 13:00:15 |
#include<iostream> #include<algorithm> #include<string> using namespace std; int main(){ int n,m; string s,ans; cin>>n>>s>>m; int x=0; for(int i=0;i<s.size();i++){ int t=s[i]>='0'&&s[i]<='9'? s[i]-'0':s[i]-'A'+10; x=x*n+t; } while(x>0){ int t=x%m; if(t<10) ans.push_back('0'+t); else ans.push_back('A'+t-10); x/=m; } reverse(ans.begin(),ans.end()); cout<<ans<<endl; return 0; }