NYOJ528 Finding the ball number (three) bit arithmetic

The problem with bit arithmetic is very simple, the premise is to first familiar with the bit operation, here to use an XOR operation, that is, ^ this symbol, his operation rule is: the same is 0, the difference is 1. After knowing this, it is easy to think of the same two number of different or after 0, so the following is a key step, but also I think for a long time did not think of a step, is to put all the numbers are different or once again, then the last remaining must be that one, there is a point to note that any number and 0 of the difference or operation is still himself Here's the code:

1#include <iostream>2#include <cstdio>3 4 using namespacestd;5 6 intMain ()7 {8     intN;9      while(~SCANF ("%d", &N))Ten     { One         intAns =0;//Initialize Condition A         intT; -          for(inti =0; I < n; i++) -         { thescanf"%d", &t); -Ans ^= t;//XOR or Operation -         } -printf"%d\n", ans); +  -     } +     return 0; A}

There is also a code to look at the STL from the Internet as follows:

#include <iostream>#include<cstdio>#include<Set>using namespacestd;intMain () {intN;  while(~SCANF ("%d", &N)) {Set<int>S; intT;  for(inti =0; I < n; i++) {scanf ("%d", &t); if(S.find (t) = =S.end ())            S.insert (t); Elses.erase (t); } printf ("%d\n", *S.begin ()); }    return 0;}

Although the STL code time is relatively long, but this is also a good method, flexible use of STL is very necessary for ACM!

NYOJ528 Finding the ball number (three) bit arithmetic