FCF2P5 - Editorial

Practice

Contest

Author: Pritish Priyatosh Nayak

Tester: Sibasish Ghosh

Editorialist: Pritish Priyatosh Nayak

Simple

PREREQUISITES:

Observation, Dynamic programming

PROBLEM:

You’re given a permutation of 1 to N.
In the permutation the position having value x is linked with the position having value x+1.
And moving from one position A to another position B takes |A-B| units of time.
Process queries which ask you to tell the minimum time needed to go from position Src to position Dest.

EXPLANATION:

Keep the position having value 1 as reference.
Let t[i] denote the time needed to go from position having value 1 to position having value i.
And then time needed to go from Src to Dest is basically abs(t[ value at Src ] - t[ value at Dest ]) .

See setter solution for implementation details.

SOLUTIONS:

Setter's/Editorialist's Solution
``````#include<bits/stdc++.h>
#define int long long
#define rep(i,x,y) for(int i=x;i<y;i++)
#define repr(i,x,y) for(int i=x;i>=y;i--)
#define mod 1000000007
using namespace std;

void Onigiri()
{
int n,q;
cin>>n>>q;
int a[n];
int pos[n];
rep(i,0,n)
{cin>>a[i];a[i]--;pos[a[i]]=i;}

int t[n];
t[0]=0;
rep(i,1,n)
{
t[i]=t[i-1]+abs(pos[i-1]-pos[i]);
}

while(q--)
{
int x,y;
cin>>x>>y;
x--,y--;
cout<<abs(t[a[x]]-t[a[y]])<<'\n';
}
}
signed main()
{
ios_base::sync_with_stdio(false);cin.tie(NULL);
#ifdef Zoro
freopen("sampletest", "r", stdin);
freopen("sampleout", "w", stdout);
#endif

int t=1;
//cin>>t;

while(t--)
{Onigiri();cout<<"\n";}

cerr<<"\n"<<(float)clock()/CLOCKS_PER_SEC*1000<<" ms"<<endl;
return 0;
}
``````
Tester's Solution
``````      #include<bits/stdc++.h>
#define int long long
using namespace std;
int32_t main()
{
ios_base::sync_with_stdio(false);cin.tie(NULL);
int n,q;
cin>>n>>q;

int a[n];
int pos[n];
for (int i = 0; i < n; ++i)
{
cin>>a[i];
a[i]--;
pos[a[i]]=i;
}
int d[n];
d[0]=0;
for (int i=1; i < n; ++i )
{
d[i]=d[i-1]+abs(pos[i-1]-pos[i]);
}

while(q--)
{
int src,dest;
cin>>src>>dest;
src--,dest--;
cout<<(int)(abs(d[a[src]]-d[a[dest]]))<<'\n';
}

cerr<<"\n"<<(float)clock()/CLOCKS_PER_SEC*1000<<" ms"<<endl;
return 0;
}

``````