标签:acm c++ 杭电 算法 编程
Find your present!
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3096 Accepted Submission(s): 2051
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
Sample Output
继续无聊中………… 清明早上竟然 下起了大学 ,天晴那么好 我也真服了东北的气候。网上买了两件衣服感觉还不错^^
明天又该回复上课了!都他妈学不会了
#include<iostream>
using namespace std;
int main()
{
int n;
int ls[201];
while(cin>>n,n)
{
int count[201]={0};
for(int i=0;i<n;i++)
cin>>ls[i];
for(int k=0;k<n;k++)
for(int j=0;j<n;j++)
if(ls[k]==ls[j])
count[k]++;
for(int m=0;m<n;m++)
if(count[m]==1)
cout<<ls[m]<<endl;
}
return 0;
}
杭电 HDU ACM 1563 Find your present!
标签:acm c++ 杭电 算法 编程
原文地址:http://blog.csdn.net/lsgqjh/article/details/44901733