B

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 that i = j.
Example
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

怎样讲呢，还是一道“水题”。这个题有想法就好不用 那么多的极限情况，错在了一个“转换”，自己当时只是考虑到在第一个0之前，i-th是FirstZero-pos减去当前位置，没能够考虑到转换之后，这种情况又是一种常态，而不是在“转换”的那一次而已，还是“嫩”啊。

#include<stdio.h>#include<algorithm>using namespace std;typedef long long ll;int main(){    ll a[200010];    int pos_zero[200010],i,n,j=0,len,flag=0;    scanf("%d",&n);    for(i=0;i<n;i++)    {        getchar();        scanf("%lld",&a[i]);        if(a[i]==0)        {            pos_zero[j]=i;            j++;        }    }    len=--j;    j=0;    for(i=0;i<n;i++)    {        if(i<=pos_zero[j]&&j<=len)        {            a[i]=pos_zero[j]-i;            if(flag)            {                a[i]=min(i-pos_zero[j-1],pos_zero[j]-i);            }        }        else if(i>pos_zero[j]&&(j+1)<=len)        {           j++;           flag=1;           a[i]=min(i-pos_zero[j-1],pos_zero[j]-i);        }        else if(i>pos_zero[j]&&(j+1)>len)        {            a[i]=i-pos_zero[len];        }    }    for(i=0;i<n-1;i++)    {        printf("%lld ",a[i]);    }    printf("%lld",a[i]);    return 0;}