1141

最新推荐文章于 2021-01-20 23:01:41 发布

最新推荐文章于 2021-01-20 23:01:41 发布 · 447 阅读

本文介绍了一个算法问题：在一串数字中找到唯一出现奇数次的数。通过示例讲解了如何利用排序和计数的方法来解决这个问题，并提供了一段C语言实现的代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述

There is a sequence S consisting of N numbers, where N is an odd number. All numbers in this sequence appear even number times, except one showing up odd number times, which we call unique number. For example, in the sequence {3, 4, 4, 9, 3, 5, 3, 3, 9}, the number 5 is the unique number. Now you are required to work out the unique number for given sequence.

输入

Input file consists of several cases.The first line of each case contains a number N. Each of the next N lines contains only one number Si. We guarantee that the number N is no more than 100,000, and Si is no more than 2^31.

输出

For each case, just return the unique number in one line.

样例输入

样例输出

提示

Try O(n) algorithm.

来源

2012.03

#include<stdio.h>
#include<string.h>
int comp(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}
int main()
{  int i,n;
   while(scanf("%d",&n)!=EOF)
  {
   int num=0;
   int a[n],m;
   for(i=0;i<n;i++)
   scanf("%ld",&a[i]);
   qsort(a,n,sizeof(a[0]),comp);
for (i=0;i<n;i++){
if (i==0||a[i]!=a[i-1]) {//第一个和每一个新出现的第一个
if (num%2) {
printf("%d\n",a[i-1]);
break;
}
num=1;//每一个新出现数的第一个记为1
}
else num++;
}}
     return 0;
}