hdu-1563 Find your present!
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关于寻找数组中的特别数字的小技巧
Find your present!
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2335 Accepted Submission(s): 1524
Problem 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.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<=200, and n is odd) at first. Following that, n positive integers will be given in a line. 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<=200, and n is odd) at first. Following that, n positive integers will be given in a line. 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
32
题目大意:找出一串数字中比较特别的数字。
解题思路:这个题目中可以用到一个小技巧,就是用异或运算,这样能快速找出来那个特别的数字。因为异或运算是“同0异1”。a^a=0 a^0=a 而且异或运算具有交换性和结合性
#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>using namespace std;int a[220];int main(){ int n; while(scanf("%d",&n)&&n) { int ans=0; for(int i=0;i<n;i++) { scanf("%d",&a[i]); ans^=a[i]; } printf("%d\n",ans); } return 0;}
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