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  1. // c++ program for the above approach
  2. #include <bits/stdc++.h>
  3. using namespace std;
  4.  
  5. void solve(int arr[], int n){
  6. 	int x=0,y=0;
  7. 	int xor1=0;
  8. 	for(int i=0;i<n;i++){
  9. 		xor1=arr[i]^xor1;
  10. 	}
  11. 	for(int i=0;i<n;i++){
  12. 		xor1= xor1^i;
  13. 	}
  14. 	int set_bit_xor = xor1& ~(xor1-1);
  15.  
  16. 	for(int i=0;i<n;i++){
  17. 		if (arr[i] & set_bit_xor)  
  18. 		{ x ^= arr[i]; }  
  19.         else              
  20.         { y ^= arr[i]; } 
  21.  
  22.         // Decide whether (i+1) is in first set 
  23.         // or second
  24.         if ((i+1) & set_bit_xor)  
  25.         { x ^= (i + 1); }
  26.         else                      
  27.         { y ^= (i + 1); }
  28. 	}
  29. 	cout << "Missing and Repeating (In any order) " 
  30.          << x << " " << y;
  31. }
  32.  
  33. int main() {
  34. 	int n;
  35. 	cin>>n;
  36. 	int arr[n];
  37. 	for(int i=0;i<n;i++)
  38. 	cin>>arr[i];
  39. 	solve(arr,n);
  40. 	return 0;
  41. }
Success #stdin #stdout 0.01s 5284KB
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stdin
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5
1 2 3 5 1
stdout
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Missing and Repeating (In any order) 0 5
https://ideone.com/XmQY6w
language:
C++ (gcc 8.3)
created:
8 days ago
visibility:
public

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