3 条题解

  • 1
    @ 2025-3-4 16:02:41
    #include <bits/stdc++.h>
    
    using namespace std;
    
    int main()
    {
      int n,m;
      cin>>n>>m;
      vector<pair<int,string>> toys(n);
      for(int i = 0;i<n;i++)
      {
        cin>>toys[i].first>>toys[i].second;
      }
      int pos = 0;
      for(int i = 0;i<m;i++)
      {
        int dir,steps;
        cin>>dir>>steps;
        if(dir == 0)
        {
          if(toys[pos].first == 0)
          {
            pos = (pos-steps+n)%n;
          }
          else {
            pos = (pos + steps)%n;
          }
        }else {
          if(toys[pos].first == 0)
          {
            pos = (pos + steps) % n;
          }
           else {
            pos = (pos-steps+n)%n;
           }
        }
      }
      cout << toys[pos].second << endl;
      return 0;
    }
    
    • 0
      @ 2025-4-26 11:53:00

      #include<bits/stdc++.h> using namespace std; int main(){ int n,m,wz=1,zl,wy; cin>>n>>m; int cx[100005]; string mz[100005]; for(int i=1;i<=n;i++){ cin>>cx[i]>>mz[i]; } for(int i=1;i<=m;i++){ cin>>zl>>wy; if(cx[wz]0){ if(zl0){ wz-=wy; } else{ wz+=wy; } } else { if(zl==0){ wz+=wy; } else{ wz-=wy; } } if(wz<1){ wz+=n; } if(wz>n){ wz-=n; } } cout<<mz[wz]; return 0; }

      • 0
        @ 2025-4-26 11:36:06

        #include <bits/stdc++.h>

        using namespace std;

        int main() { int n,m; cin>>n>>m; vector<pair<int,string>> toys(n); for(int i = 0;i<n;i++) { cin>>toys[i].first>>toys[i].second; } int pos = 0; for(int i = 0;i<m;i++) { int dir,steps; cin>>dir>>steps; if(dir == 0) { if(toys[pos].first == 0) { pos = (pos-steps+n)%n; } else { pos = (pos + steps)%n; } }else { if(toys[pos].first == 0) { pos = (pos + steps) % n; } else { pos = (pos-steps+n)%n; } } } cout << toys[pos].second << endl; return 0; }

        • @ 2025-4-26 11:53:27

          #include<bits/stdc++.h> using namespace std; int main(){ int n,m,wz=1,zl,wy; cin>>n>>m; int cx[100005]; string mz[100005]; for(int i=1;i<=n;i++){ cin>>cx[i]>>mz[i]; } for(int i=1;i<=m;i++){ cin>>zl>>wy; if(cx[wz]0){ if(zl0){ wz-=wy; } else{ wz+=wy; } } else { if(zl==0){ wz+=wy; } else{ wz-=wy; } } if(wz<1){ wz+=n; } if(wz>n){ wz-=n; } } cout<<mz[wz]; return 0; }

      • 1

      信息

      ID
      5621
      时间
      1000ms
      内存
      500MiB
      难度
      2
      标签
      递交数
      110
      已通过
      41
      上传者