Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4044 Accepted Submission(s): 2671 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. Output For each case, output an integer in a line, which is the card number of your present. Sample Input 5 1 1 3 2 2 3 1 2 1 0 Sample Output 3 2 Author 8600 Source Happy 2007 简单地说题意就是让你从一堆数里找到只出现一次的数字。 #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace std; int main() { int n; int a[205]; while(cin>>n,n!=0) { int i; for(i=0;i<n;i++) cin>>a[i]; sort(a,a+n);//优化在这里，而且我的循环也只有一层 if(a[0]!=a[1]) cout<<a[0]<<endl; for(i=1;i<n-1;i++) { if(a[i]!=a[i-1]&&a[i]!=a[i+1]) cout<<a[i]<<endl; } if(a[n-1]!=a[n-2]) { cout<<a[n-1]<<endl; } } return 0; }
