Distances to Zero

2024-01-02 13:08

文章标签 zero distances

本文主要是介绍Distances to Zero，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

You are given the array of integer numbers a₀, a₁, ..., a_n - 1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array.

Input

The first line contains integer n (1 ≤ n ≤ 2·10⁵) — length of the array a. The second line contains integer elements of the array separated by single spaces ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

Print the sequence d₀, d₁, ..., d_n - 1, where d_i is the difference of indices between i and nearest j such that a_j = 0. It is possible that i = j.

Examples

Input

9
2 1 0 3 0 0 3 2 4

Output

2 1 0 1 0 0 1 2 3

Input

5
0 1 2 3 4

Output

0 1 2 3 4

Input

7
5 6 0 1 -2 3 4

Output

2 1 0 1 2 3 4

给出一个数列，求每个数距离数列中0的最近距离；

思路：记录每个0的位置，然后从前往后遍历一遍求距离0的距离，如果某个数在两个零之间，则取较小值；

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>using namespace std;int num[200005];
int zero[200005];
int main(){int n;cin >> n;int k=0;for(int i=1; i<=n; i++){cin >> num[i];if(num[i]==0) zero[++k]=i;}int dis=0;int cnt=k;k=1;zero[0]=0;for(int i=1; i<=n; i++){dis=abs(i-zero[k]);if(dis==0){k++;}if(k>1 && zero[k]!=0 && k<=cnt) dis=min(dis, abs(i-zero[k-1]));   //如果位于两个0之间，取较小值；if(k>cnt) dis=abs(i-zero[cnt]);//cout << "k:" << k << endl;cout << dis;if(i==n) cout << endl;else cout << ' ';}
return 0;
}

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