HDU - 1029 Ignatius and the Princess IV（DP）

最新推荐文章于 2020-02-07 11:05:28 发布

Wabrush

最新推荐文章于 2020-02-07 11:05:28 发布

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分类专栏：动态规划文章标签： dp HDU

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本文介绍了一个有趣的问题：从一组奇数个元素中找到至少出现（n+1）/2次的特殊整数，并提供了一种O(n)时间复杂度的算法实现。

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 "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

挺有意思的一道题，给你n（n为奇数）个数，让你找出出现次数不小于（n+1)/ 2的数。
题目不难，但听说这题有一种时间复杂度为O(n)的算法，但还是被我想出来了！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#define N 1000100
#define INF 0x3f3f3f3f
using namespace std;
int main()
{
    int n;
    while(cin >> n)
    {
        int a;
        int pos = 0, num_pos = 0;//pos为当前出现次数对多的数字， num_pos为其出现次数
        int t = 0, num_t = 0;//t为当前数字， num_t为当前数字出现的次数
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a);
            if(a == t)
            {
                num_t++;
                if(num_t > num_pos)//若当前数字的出现次数大于上一个出现次数最多的数字的出现次数，则迭代
                {
                    pos = t;
                    num_pos = num_t;
                }
            }
            else if(a == pos)
            {
                num_pos++;
            }
            else
            {
                t = a;
                num_t = 0;
            }
        }
        printf("%d\n", pos);
    }
    return 0;
}