A. Be Positive

A. Be Positive

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array of nn integers: a1,a2,…,ana1,a2,…,an. Your task is to find some non-zero integer dd (−103≤d≤103−103≤d≤103) such that, after each number in the array is divided by dd, the number of positive numbers that are presented in the array is greater than or equal to half of the array size (i.e., at least ⌈n2⌉⌈n2⌉). Note that those positive numbers do not need to be an integer (e.g., a 2.52.5 counts as a positive number). If there are multiple values of dd that satisfy the condition, you may print any of them. In case that there is no such dd, print a single integer 00.

Recall that ⌈x⌉⌈x⌉ represents the smallest integer that is not less than xx and that zero (00) is neither positive nor negative.

Input

The first line contains one integer nn (1≤n≤1001≤n≤100) — the number of elements in the array.

The second line contains nn space-separated integers a1,a2,…,ana1,a2,…,an (−103≤ai≤103−103≤ai≤103).

Output

Print one integer dd (−103≤d≤103−103≤d≤103 and d≠0d≠0) that satisfies the given condition. If there are multiple values of dd that satisfy the condition, you may print any of them. In case that there is no such dd, print a single integer 00.

Examples

input

5
10 0 -7 2 6

output

4

input

7
0 0 1 -1 0 0 2

output

0

Note

In the first sample, n=5n=5, so we need at least ⌈52⌉=3⌈52⌉=3 positive numbers after division. If d=4d=4, the array after division is [2.5,0,−1.75,0.5,1.5][2.5,0,−1.75,0.5,1.5], in which there are 33 positive numbers (namely: 2.52.5, 0.50.5, and 1.51.5).

In the second sample, there is no valid dd, so 00 should be printed.

题解：计算出正数和负数的个数，如果正数大于（n+1）/2的话，就是随便输出一个正数，如果负数大于（n+1）/2的话，就随便输出一个负数，否则输出0.

#include<iostream>
#include<cstdio>

using namespace std;

int main()
{
    int n,a[1005],cnt=0,cnt2=0;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        scanf("%d",a+i);
        if(a[i]>0)        //找出正数的个数
            cnt++;    
        else if(a[i]<0)        //找出负数的个数
            cnt2++;

    }    
    if(cnt>=(n+1)/2)
        printf("1\n");
    else if(cnt2>=(n+1)/2)
        printf("-1\n");
    else
        printf("0\n");
    return 0;
}

 

转载于:https://www.cnblogs.com/buhuiflydepig/p/10638523.html