杭电ACM题1001的超时问题

最新推荐文章于 2024-02-25 18:23:11 发布

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最新推荐文章于 2024-02-25 18:23:11 发布

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分类专栏：杭电ACM 文章标签： input

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本文介绍了一个ACM竞赛问题，任务是在一系列整数中找到至少出现(N+1)/2次的特殊整数，并给出了使用C++解决该问题的有效代码示例。

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Ignatius and the Princess IV

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32767K (Java/Other)

Total Submission(s) : 65 Accepted Submission(s) : 17

Problem Description

"OK, you are not too bad, em... But you can never pass the next test." feng5166 says.

"I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says.

"But what is the characteristic of the special integer?" Ignatius asks.

"The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says.

Can you find the special integer for Ignatius?

Input

The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file.

Output

For each test case, you have to output only one line which contains the special number you have found.

Sample Input

5
1 3 2 3 3
11
1 1 1 1 1 5 5 5 5 5 5
7
1 1 1 1 1 1 1

Sample Output

3
5
1

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

简单题，做的时候出现了超时问题，问题是由于cin造成的，用cin输入num会出现超时，但通过scanf函数接受num时可以AC，这个不明白，刚开始学C++，对它的输入输出不是很了解。

//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

代码：

/***********************
* 程序名:Ignatius and the Princess IV.cpp
* 功能:ACM
************************/
#include <iostream>
#include <Cstdio>

using namespace std;

int main()
{
    int N, num, temp, i;
    while(scanf("%d", &N) != EOF) {
        int count[32768] = {0};
        for(i = 0; i < N; i++) {
            scanf("%d", &num);
            count[num]++;
            if(count[num] >= (N+1)/2) {
                temp = num;
            }
        }
        cout<<temp<<endl;
    }
    return 0;
}