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HDU2095 find your present (2)【水题】

2024-11-06 07:27:39   204人阅读

find your present (2)


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17030    Accepted Submission(s): 6425

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, and you can assume that only one number appear odd times.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<1000000, and n is odd) at first. Following that, n positive integers will be given in a line, all integers will smaller than 2^31. 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
Hint
use scanf to avoid Time Limit Exceeded

Author
8600
 
Source

HDU 2007-Spring Programming Contest - Warm Up （1）


题目大意：给你N个数(N为奇数)，找到N个数里边有奇数个的那个数。

思路：排序后统计每个数的个数，输出奇数的那个数。


#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

int a[1000010];
int main()
{
    int N;
    while(scanf("%d",&N) && N!=0)
    {
        memset(a,0,sizeof(a));
        for(int i = 0; i < N; ++i)
        {
            scanf("%d",&a[i]);
        }
        if(N==1)
        {
            printf("%d\n",a[0]);
            continue;
        }
        int i,num = 1;
        sort(a,a+N);
        for(i = 1; i < N; i++)
        {
            if(a[i]==a[i-1])
                num++;
            else
            {
                if(num&1)
                {
                    printf("%d\n",a[i-1]);
                    break;
                }
                else
                    num = 1;
            }
        }
        if(i==N)
            printf("%d\n",a[N-1]);
    }

    return 0;
}


HDU2095 find your present (2)【水题】


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