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  1. #include<bits/stdc++.h>
  2. using namespace std;
  3.  
  4. const int MaxN=1e5;
  5. const int MaxA=1e9;
  6. int N, M;
  7. int A[MaxN];
  8. int B[MaxN];
  9.  
  10. int main(){
  11. 	// input
  12. 	cin>>N>>M;
  13. 	for(int n=0; n<N; n+=1)
  14. 		cin>>A[n];
  15. 	for(int m=0; m<M; m+=1)
  16. 		cin>>B[m];
  17. 	// 
  18. 	sort(A,A+N);
  19. 	sort(B,B+M);
  20. 	long W=0;
  21. 	for(int m=0, n=0; n<N; n+=1){
  22. 		while(m<M && B[m]<A[n])
  23. 			m+=1;
  24. 		W+=m;
  25. 	}
  26. 	long L=0;
  27. 	for(int n=0, m=0; m<M; m+=1){
  28. 		while(n<N && A[n]<B[m])
  29. 			n+=1;
  30. 		L+=n;
  31. 	}
  32.  
  33. 	int l=M, y=0;
  34. 	for(int w=0, n=0; n<N; n+=1){
  35. 		while(w<M && B[w] <A[n])
  36. 			w+=1;
  37. 		while(y<M && B[y]<=A[n]){
  38. 			y+=1;
  39. 			l-=1;
  40. 		}
  41. 		if( w>=l && w+W>l+L ){
  42. 			cout<<A[n];
  43. 			return 0;
  44. 		}
  45. 	}
  46. 	cout<<-1;
  47. }
Success #stdin #stdout 0s 5304KB
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https://ideone.com/qLtF6b
e864. Q3-2 循環小數
language:
C++ (gcc 8.3)
created:
11 days ago
visibility:
public

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