CF#803 B. Distances to Zero（水题）

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 numbers a0, a1, ..., an - 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 ≤ ai ≤ 109).

Output

Print the sequence d0, d1, ..., dn - 1, where di is the difference of indices between i and nearest j such that aj = 0. It is possible thati = j.

Examples

input

92 1 0 3 0 0 3 2 4

output

2 1 0 1 0 0 1 2 3 

input

50 1 2 3 4

output

0 1 2 3 4 

input

75 6 0 1 -2 3 4

output

2 1 0 1 2 3 4 

题意：计算每个数到最近的0的距离

#include <bits/stdc++.h>using namespace std;const int N = 2e5 + 10;int n;int a[N],b[N],c[N];int main(){    while(scanf("%d", &n) == 1)    {        int x = 0;        for(int i = 0; i < n; i++)        {            scanf("%d", &a[i]);            if(!a[i]) b[x++] = i;        }        int k = 0;        for(int i = 0; i < b[x - 1]; i++)        {            for(int j = b[i]; j <= b[i + 1]; j++)            {                a[j] = max(0,min(j - b[i], b[i + 1] - j));            }        }        for(int i = 0; i < b[0]; i++)        {            a[i] = b[0] - i;        }        for(int i = b[x - 1]; i < n; i++)        {            a[i] = i - b[x - 1];        }        for(int i = 0; i < n; i++)        {            printf("%d%c", a[i], i < n - 1 ? ' ' : '\n');        }    }}