【HUD 1029】Ignatius and the Princess IV（思维-水题）

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Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 33888    Accepted Submission(s): 14702

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

51 3 2 3 3111 1 1 1 1 5 5 5 5 5 571 1 1 1 1 1 1

 

Sample Output

351

 

Author

Ignatius.L

 

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题意：找出那个出现次数超过（N+1）/2的数字；

题解：可以直接暴力解，如果先进行sort快排，因为N是奇数，出现（N+1）/2次的数字无论在哪个位置都一定会出现在num【（n+1）/2】这个位置上，快排后直接输出num【（n+1）/2】但是有可能会超时；

代码：（下标表示法：（暴力解法）就是每个输入的数直接作为数组下标，数组元素就是这个数字出现的次数）

#include<stdio.h>#include<iostream>#include<cstring>using namespace std;const int MAX=1e6+10;int sum[MAX];int main(){int n;while(~scanf("%d",&n)){memset(sum,0,sizeof(sum));int temp=(n+1)/2;int num;int ans;for(int i=0;i<n;i++){scanf("%d",&num);sum[num]++;if(sum[num]>=temp)ans=num;}cout<<ans<<endl;}}