HDU 1029.Ignatius and the Princess IV【数据弱鸡】【咦，水题！】【8月20】

Ignatius and the Princess IV

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

给你n个数，输出其中出现次数大于等于(n+1)/2的数。直接做就能过，这题数据太弱了，不然肯定超时。代码如下：

#include<cstdio>#include<cstring>int f[1000100];int main(){    int n,ans,x;    while(scanf("%d",&n)==1){        memset(f,0,sizeof(f));        for(int i=0;i<n;i++){            scanf("%d",&x);            f[x]++;            if(f[x]>=(n+1)/2)                ans=x;        }        printf("%d\n",ans);    }    return 0;}

但是上面这种方法用的空间有点大，尝试用map做。代码如下：

#include<cstdio>#include<map>using namespace std;int main(){    int n,ans,x;    map<int,int> f;    while(scanf("%d",&n)==1){        f.clear();        for(int i=0;i<n;i++){            scanf("%d",&x);            f[x]++;            if(f[x]>=(n+1)/2)                ans=x;        }        printf("%d\n",ans);    }    return 0;}

时间效率不高····充分说明了这个题的数据弱！