BestCoder Round #50 Distribution money 即 hdu5364 (排序)

Distribution money

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

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Total Submission(s): 379    Accepted Submission(s): 227

Problem Description

AFA want to distribution her money to somebody.She divide her money into n same parts.One who want to get the money can get more than one part.But if one man's money is more than the sum of all others'.He shoule be punished.Each one who get a part of money would write down his ID on that part.

Input

There are multiply cases.
For each case,there is a single integer n(1<=n<=1000) in first line.
In second line,there are n integer a1,a2...an(0<=ai<10000)ai is the the ith man's ID.

Output

Output ID of the man who should be punished.
If nobody should be punished,output -1.

Sample Input

3 1 1 2 4 2 1 4 3

Sample Output

1 -1

Source

​​BestCoder Round #50 (div.2)​​

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解析：题意就是对于长度为n的序列，判断其中是否有一个数字的出现次数超过n/2次。若有，输出这个数字；若无，输出-1。

          直接读入之后排序，然后扫描判断时候存在一段连续的相等数字个数大于n/2。

代码：

#include<cstdio>
#include<algorithm>
#define maxn 1000
using namespace std;

int a[maxn+20];

int main()
{
  int i,j,k,n;
  while(scanf("%d",&n)!=EOF)
    {
      for(i=1;i<=n;i++)scanf("%d",&a[i]);
      sort(a+1,a+n+1);
      k=n/2,j=1;
      for(i=2;i<=n;i++)
        {
          if(a[i]==a[i-1]){j++;continue;}
          if(j>k){i--;break;}
          j=1;
    }
    if(j>k)printf("%d\n",a[i]);
    else printf("-1\n");
  }
  return 0;
}