B. Distances to Zero time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

You are given the array of integer numbersa₀, 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 integern(1 ≤ n ≤ 2·10⁵) — length of the arraya. The second line contains integer elements of the array separated by single spaces ( - 10⁹

 ≤ a_i ≤ 10⁹).

Output

Print the sequenced₀, d₁, ..., d_n - 1, whered_iis the difference of indices betweeniand nearestjsuch thata_j = 0. It is possible thati = 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 

題目大意：n個元素，求每個數到離它最近的0的距離 思路：模擬瞎暴力，記錄相鄰兩個0的位置，依次做比較

#include <bits/stdc++.h>
#define manx 200005
#define INF 0x3f3f3f3f
using namespace std;

int main()
{
    int a[manx],ans[manx],n;
    while(~scanf("%d",&n)){
        memset(a,0,sizeof(a));
        memset(ans,0,sizeof(ans));
        int k[2],t=0;  //k記錄相鄰兩個0的位置
        k[0]=k[1]=INF;
        for (int i=0; i<n; i++)
            scanf("%d",&a[i]);
        for (int i=0; i<n+1; i++){
            if (!a[i]){
                ans[i]=0;
                if (i!=n) k[t%2]=i;
                int f;
                if (k[(t+1)%2]==INF) f=0;
                else f=k[(t+1)%2];
                for (int j=f; j<i; j++)
                    ans[j]=min(abs(k[0]-j),abs(k[1]-j));
                t++;
            }
        }
        for (int i=0; i<n; i++)
            printf("%d ",ans[i]);
        printf("\n");
    }
    return 0;
}