Write and Erase

标签：following   偶数   hat   ase   format   return   process   turn   namespace   

You are playing the following game with Joisino.

• Initially, you have a blank sheet of paper.
• Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times.
• Then, you are asked a question: How many numbers are written on the sheet now?

The numbers announced by Joisino are given as A₁,…,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?

Constraints

• 1≤N≤100000
• 1≤A_i≤1000000000(=10⁹)
• All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A1
:
AN

Output

Print how many numbers will be written on the sheet at the end of the game.

Sample Input 1

3
6
2
6

Sample Output 1

1

The game proceeds as follows:

• 6 is not written on the sheet, so write 6.

• 2 is not written on the sheet, so write 2.

• 6 is written on the sheet, so erase 6.

Thus, the sheet contains only 2 in the end. The answer is 1.

Sample Input 2

4
2
5
5
2

Sample Output 2

0

It is possible that no number is written on the sheet in the end.

Sample Input 3

6
12
22
16
22
18
12

Sample Output 3

2



该题就是判断一个数出现的次数，偶数为0，奇数+1；

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
long long a[1000010];
int main()
{
    int n,x=1,ans=0;//x表示一个数出现的次数，赋值1 
    cin>>n;
    for(int i=0;i<n;i++)
       cin>>a[i];
    sort(a,a+n);//先排序，更方便判断奇偶 
    for(int i=1;i<n;i++)
    {
        if(a[i]==a[i-1])//出现相同的数，x+1然后mod2 
          x=(x+1)%2;
        else
        {
            ans+=x;//单独的一个数，ans+1，x重新赋值为1，进行循环 
            x=1;
        }      
    }
    if(x%2)//再次判断 ，若不为0，则ans+1 
      ans++;
    cout<<ans<<endl;
    return 0;
} 

Write and Erase

标签：following   偶数   hat   ase   format   return   process   turn   namespace   

原文地址：https://www.cnblogs.com/ylrwj/p/10792504.html