时间复杂度：O(1)O(1)O(1)

#include<iostream> using namespace std; int main(){ long long x1,x2,y1,y2; cin>>x1>>x2; cin>>y1>>y2; if(x1y1<=-2147483647||x2y2>=2147483647){ cout << "long long int"; }else{ cout << "int"; } return 0; }

<协助人员DeepSeek>

开long long有时是Accepted与Unaccepted之间的永恒主题。