K8S 中通過 service 訪問外部的資源
阿新 • • 發佈:2020-12-29
#include<iostream> #include <algorithm> using namespace std; struct node{ int first; int second; }; node x[500005]; node y[500005]; int main() { ios::sync_with_stdio(false); cin.tie(0); int t; cin>>t; while(t--){ int n,r,s=-1,count=0; cin>>n; for(int i=1;i<=n;i++){ cin>>r; if(s==r){ //保證了x中數字的不重複 if(r!=y[count].first){ y[++count].first = r; y[++count].second = 0; } // break; }else{ s=r; x[i].first = r; x[i].second = 0; } } for(int i=1;i<=count;i++){ y[i].first+=1; y[i].second = 1;//表示被修改過了 } while (true){ } } return 0; }
#include<iostream> #include <algorithm> #include <string> using namespace std; int x[500005]; int main() { ios::sync_with_stdio(false); cin.tie(0); int t; cin>>t; // t = 1; while(t--){ string s; int count = 0; cin>>s; int len = s.length(); for(int i=0;i+3<len;i++){ if(s[i]==s[i+1]&&s[i+1]==s[i+2]&&s[i+2]==s[i+3]&&x[i]+x[i+1]+x[i+2]+x[i+2]==0){ x[i+1]=x[i+2]=1; count += 2; i+=4; } } // cout<<"len=="<<len<<endl; for(int i=1;i<=len-1;i++){ if(s[i] == s[i-1] && x[i]+x[i-1] == 0){ if((i>=2&&s[i-2]==s[i]&&x[i-2]+x[i]==0)||(i+2<len&&s[i+2]==s[i]&&x[i+2]+x[i]==0)){ x[i]=1; } else{ x[i-1]=1; } count++; } } /* for(int i =0;i<len;i++){ cout<<x[i]<<" "; } cout<<endl;*/ for (int j = 1; j <= len-2 ; ++j) { if(s[j-1] == s[j+1] && x[j-1]+x[j+1] == 0){ x[j-1] = 1; count++; } } for(int i = 0;i<len;i++){ x[i] = 0; } cout<<count<<endl; } return 0; }