HDU -1029 Ignatius and the Princess IV

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最新推荐文章于 2023-01-19 14:24:41 发布

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分类专栏： 2018暑假ACM集训 # ACM_动态规划、贪心

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本文介绍了一个算法问题，即在一组由奇数个整数组成的数据中，如何快速找到那个出现次数超过总数一半的特殊整数。文章通过示例解释了解决方案，并提供了一段C++代码实现。

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Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)	Memory Limit: 65536/32767 K (Java/Others)

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

题意：找到数量大于n/2个的内个数

可以一个一个抵消，这个跟前一个不一样就抵消掉，最后剩下的就是答案。第二个样例，前面累积了5个1，后面5个5跟它抵消，就剩个5了。

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

int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        int t=0, x, ans;//ans存当前的答案，t存ans出现的次数，后面的跟ans一样++不一样--
        while(n--)
        {
            scanf("%d", &x);
            if(t<=0)
                ans=x;
            x==ans ? t++:t--;
        }
        printf("%d\n", ans);
    }
}