【CodeForces 803B】Distances to Zero（模擬）



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, ..., 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 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;
}