Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.
The first line contains an integer n (1 ≤ n ≤ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 ≤ l < r ≤ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.
8
1 6 5 4 3 2 7 8
2 6
4
2 3 4 1
0 0
4
1 2 3 4
0 0
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n,a[1010];
int main()
{
int l=0,r=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
if(a[i]!=i)
{
l=i;
break;
}
for(int i=n;i>=1;i--)
if(a[i]!=i)
{
r=i;
break;
}
for(int i=l;i<r;i++)
{
if(a[i]<a[i+1])
{
printf("0 0
");
return 0;
}
}
if(l==0&&r==0)
{
printf("0 0
");
return 0;
}
printf("%d %d",l,r);
return 0;
}