46 人解决，55 人已尝试。
68 份提交通过，共有 134 份提交。
3.4 EMB 奖励。
单点时限: 2.0 sec
内存限制: 256 MB
The cows are so very silly about their dinner partners. They have organized themselves into two groups (conveniently numbered 1 and 2) that insist upon dining together in order, with group 1 at the beginning of the line and group 2 at the end. The trouble starts
when they line up at the barn to enter the feeding area.
Each cow i carries with her a small card upon which is engraved D_i (1 <= D_i <= 2) indicating her dining group membership. The entire set of N (1 <= N <= 30,000) cows has lined up for dinner but it’s easy for anyone to see that they are not grouped by their dinner-partner cards.
FJ’s job is not so difficult. He just walks down the line of cows changing their dinner partner assignment by marking out the old number and writing in a new one. By doing so, he creates groups of cows like 112222 or 111122 where the cows’ dining groups are sorted
in ascending order by their dinner cards. Rarely he might change cards so that only one group of cows is left (e.g., 1111 or 222).
FJ is just as lazy as the next fellow. He’s curious: what is the absolute minimum number of cards he must change to create a proper grouping of dining partners? He must only change card numbers and must not rearrange the cows standing in line.
Line 1: A single integer: N
Lines 2..N+1: Line i+1 describes cow i’s dining preference with a single integer: D_i
7 2 1 1 1 2 2 1 INPUT DETAILS: Seven cows; all but three of which currently prefer dining group 1.
2 OUTPUT DETAILS: Change the first and last cow's cards.
46 人解决，55 人已尝试。
68 份提交通过，共有 134 份提交。
3.4 EMB 奖励。