SRM 572 250 DIV2

菜鸟的代码：

import java.util.*;
import java.util.regex.*;
import java.text.*;
import java.math.*;


public class EasyHomework
{
    public String determineSign(int[] A)
    {
        int i,count;
        count=0;
        
        for(i=0;i<A.length;i++){
            if(A[i]<0)
                count++;
            else if(A[i]==0)
                return "ZERO";
                
        }
        
        if(count%2>0)
            return "NEGATIVE";
        else
            return "POSITIVE";
    }
    

}
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 大神的代码（JAVA）：

public class EasyHomework {
 
  public String determineSign(int[] A)
  {
    int minus=0;
    for(int i=0;i<A.length;i++)
    {
      if(A[i]<0) minus++;
      else if(A[i]==0) return "ZERO";
    }
    if(minus%2==0) return "POSITIVE";
    else return "NEGATIVE";
  }
 
}

大神的代码（C++）：

#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
 
#include <algorithm>
#include <fstream>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
 
using namespace std;
using namespace __gnu_cxx;
 
typedef long long int64;
 
#define FORIT(it, v) for (typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
 
class EasyHomework {
public:
    string determineSign( vector <int> A ) {
        int minus = 0;
        FORIT(it, A) {
            if (*it == 0)
                return "ZERO";
            if (*it < 0)
                ++minus;
        }
 
        if (minus % 2 == 0)
            return "POSITIVE";
 
        return "NEGATIVE";
    }
};
 
 
 
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分析：

　　算法：

　　对比：

　　总结：嗯。。都是一样的，太简单了。