Михаил Долинский
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Мой профиль

(Analysis by Dhruv Rohatgi )
Notice that if the counter is a positive number $i$ on some day, then the counter must be $i1$ on the previous day. So we can sweep from the last day backwards in time, applying this rule to fill in missing entries. If we try to fill in an entry which was not missing, then either nothing happens (if we try to fill in some number $n$, and the entry is already $n$), or we get a contradiction (if we try to fill in a number $n$, but the entry contains a different nonnegative number).
The first entry in the log is a special case, since Farmer John already knows the cows broke out on that day. So if it is a missing entry, we can set it to $0$. If it is any positive number, then we have a contradiction.
If we ever come across a contradiction, then the log is necessarily inconsistent, so we can just output $1$. Otherwise, the log must consist of several streaks $0, 1, 2, \dots, k$ of various lengths, with possibly some $1$s between streaks  entries which we could not uniquely deduce. We know that the first streak starts on the first day.
Suppose there are $s$ streaks and $t$ entries that are still missing. Then to minimize the number of breakouts, each sequence of consecutive missing entries would be filled in so that it continues the streak preceding it (e.g. $0,1,2,3,1,1$ would be filled in as $0,1,2,3,4,5$). So the minimum number of breakouts is $s$.
To maximize the number of breakouts, the cows would have broken out on each of the $t$ days for which the entry is missing (e.g. $0,1,2,3,1,1$ would be filled in as $0,1,2,3,0,0$). So the maximum number of breakouts is $s+t$.
#include <iostream>
#include <algorithm>
using namespace std;
#define MAXN 100000
int N;
int A[MAXN];
int main()
{
cin >> N;
for(int i=0;i<N;i++)
cin >> A[i];
if(A[0] > 0)
{
cout << 1 << '\n';
return 0;
}
A[0] = 0;
int t = 1;
int req = 0;
int pos = 0;
for(int i=N1;i>=0;i)
{
if(t != 1 && A[i] != 1 && A[i] != t)
{
cout << 1 << '\n';
return 0;
}
if(t == 1)
t = A[i];
if(A[i] == 1)
A[i] = t;
if(A[i] == 0)
req++;
if(A[i] == 1)
pos++;
if(t > 1)
t;
}
cout << req << ' ' << req+pos << '\n';
}
