Algorithm

B. Spoilt Permutation
time limit per test
2 seconds
memory limit per test
256 megabytes
Problem Link- https://codeforces.com/problemset/problem/56/B
input
standard input
output
standard output

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.

Input

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.

Output

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.

Examples
input
Copy
8
1 6 5 4 3 2 7 8
output
Copy
2 6
input
Copy
4
2 3 4 1
output
Copy
0 0
input
Copy
4
1 2 3 4
output
Copy
0 0



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <cstdio>
#include <vector>

int main(){

    int n; scanf("%d", &n);
    std::vector<int> a(n + 1);
    int l(-1), r(-1);

    for(int p = 1; p <= n; p++){
        scanf("%d", &a[p]);
        if(p != a[p]){
            if(l < 0){l = p;}
            r = p;
        }
    }

    if(l < 0){puts("0 0"); return 0;}
    for(int p = l; p <= r; p++){if(a[p] != l + r - p){puts("0 0"); return 0;}}
    printf("%d %d\n", l, r>;

    return 0;
}
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Input

x
–
+
cmd
8
1 6 5 4 3 2 7 8

Output

x
–
+
cmd
2 6
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Demonstration

Codeforces Solution-B. Spoilt Permutation-Solution in C, C++, Java, Python

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