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 that i = 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
Solve:
#include<bits/stdc++.h>
using namespace std;
vector<long long>vec;
int main()
{
long long n,i,c,j,a,in;
while(cin >>n)
{
for(i=0; i<n; i++)
{
cin>>in;
vec.push_back(in);
}
if(n==1)
cout<<0;
else
for(i=0; i<n; i++)
{
c=1;
//cout<<"f;kasj";
if(i!=0)
cout<<" ";
while(c<n)
{
if(vec[i]==0)
{
cout<<0;
break;
}
else if(vec[i+c]==0&&i+c<n)
{
cout<<c;
break;
}
else if(vec[i-c]==0&&i-c>=0)
{
cout<<c;
break;
}
c++;
}
}
cout<<endl;
}
return 0;
}
এই প্রবলেম টা যদি আমরা এভাবে সল্ভ করি তাহলে time limit হওয়ার সম্ভাবনা থাকে।
সে জন্য নিচের নিয়মে করতে হবে।
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long n,i,a,mn,c,cc;
while(cin>>n)
{
vector<long long>vec,v;
cc=0;
for(i=0; i<n; i++)
{
cin>>a;
vec.push_back(a);
if(a==0)
{
v.push_back(i);
cc++;
}
}
c=0;
for(i=0; i<n; i++)
{
// if(c+1<=cc)
// v.push_back(-9);
//cout<<i<<" bbb "<<v[c]<<" nbn "<<v[c+1]<<endl;
//if(v[c+1]!=0
mn=min(abs(i-v[c]),abs(i-v[c+1]));
cout<<mn;
if(i==v[c+1])
c++;
if(i!=n-1)
cout<<" ";
}
cout<<endl;
}
return 0;
}
