find your present (2)
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find your present (2)
Time Limit: 1000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8213 Accepted Submission(s): 2224Problem Description
In the new year party, everybody will get a "special present".Now it's your turn to get your special present, a lot of presents now putting on the desk, and only one of them will be yours.Each present has a card number on it, and your present's card number will be the one that different from all the others, and you can assume that only one number appear odd times.For example, there are 5 present, and their card numbers are 1, 2, 3, 2, 1.so your present will be the one with the card number of 3, because 3 is the number that different from all the others.
Input
The input file will consist of several cases.
Each case will be presented by an integer n (1<=n<1000000, and n is odd) at first. Following that, n positive integers will be given in a line, all integers will smaller than 2^31. These numbers indicate the card numbers of the presents.n = 0 ends the input.
Each case will be presented by an integer n (1<=n<1000000, and n is odd) at first. Following that, n positive integers will be given in a line, all integers will smaller than 2^31. These numbers indicate the card numbers of the presents.n = 0 ends the input.
Output
For each case, output an integer in a line, which is the card number of your present.
Sample Input
51 1 3 2 231 2 10
Sample Output
32use scanf to avoid Time Limit ExceededHintHint
Author
8600
Source
HDU 2007-Spring Programming Contest - Warm Up (1)
Recommend
8600
经过田神的指导,发现这个题目用位运算的异或运算可以快速的求解,大家可以去了解一下,好神奇啊。
#include <iostream>#include <cstdio>using namespace std;int main(){ int x; int result; int n; while(cin>>n){ if(n==0) break; result = 0; while(n--){ cin>>x; result = result^x; } cout<<result<<endl; }}
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