Distances to Zero

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

题意:给你一串数，其中包含0，问每个数到最近的0的距离；

思路：记录每个0的位置，然后模拟就行了；

下面附上我的代码：

#include<bits/stdc++.h>
using namespace std;
int a[200005],b[200005];
int k,n;
int main()
{
	while(~scanf("%d",&n))
	{
		k=0;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			if(!a[i])
				b[k++]=i;
		}
		for(int i=0;i<n;i++)
		{
			if(i<b[0])
			{
				printf("%d%c",b[0]-i,i==n-1?'\n':' ');
				continue;
			}
			else if(i>b[k-1])
			{
				printf("%d%c",i-b[k-1],i==n-1?'\n':' ');
				continue;
			}
			int p=lower_bound(b,b+k,i)-b;
			//printf("====%d\n",p);
			printf("%d%c",min(b[p]-i,i-b[p-1]),i==n-1?'\n':' ');
		}
	}
	return 0;
 }