2019 GDUT Rating Contest II : A. Taming the Herd

题面：

A. Taming the Herd

Input file: standard input

Output file: standard output

Time limit: 1 second

Memory limit: 256 megabytes

 

Early in the morning, Farmer John woke up to the sound of splintering wood. It was the cows, and they were breaking out of the barn again!

Farmer John was sick and tired of the cows’ morning breakouts, and he decided enough was enough: it was time to get tough. He nailed to the barn wall a counter tracking the number of days since the last breakout. So if a breakout occurred in the morning, the counter would be 0 that day; if the most recent breakout was 3 days ago, the counter would read 3. Farmer John meticulously logged the counter every day.

The end of the year has come, and Farmer John is ready to do some accounting. The cows will pay, he says! But lo and behold, some entries of his log are missing!

Farmer John is confident that the he started his log on the day of a breakout. Please help him determine, out of all sequences of events consistent with the log entries that remain, the minimum and maximum number of breakouts that may have take place over the course of the logged time.

 

Input

The first line contains a single integer N (1 ≤ N ≤ 100), denoting the number of days since Farmer John started logging the cow breakout counter.

The second line contains N space-separated integers. The ith integer is either −1, indicating that the log entry for day i is missing, or a non-negative integer ai (at most 100), indicating that on day i the counter was at ai.

 

Output

If there is no sequence of events consistent with Farmer John’s partial log and his knowledge that the cows definitely broke out of the barn on the morning of day 1, output a single integer −1. Otherwise, output two space-separated integers m followed by M, where m is the minimum number of breakouts of any consistent sequence of events, and M is the maximum.

 

Example

Input

4

-1 -1 -1 1

Output

2 3

 

Note

In this example, we can deduce that a breakout had to occur on day 3. Knowing that a breakout also occurred on day 1, the only remaining bit of uncertainty is whether a breakout occurred on day 2. Hence, there were between 2 and 3 breakouts in total.

 

题目描述：

奶牛搞破坏，把奶牛棚弄坏了。从奶牛棚被破坏的那一天起，农夫就开始写日志，并且每一天都会写。这个日志记录的是：离上次奶牛棚被破坏的天数。如果当天奶牛棚被损坏，就记录为0；如果前3天奶牛棚被损坏，就记录为3。年末，农夫拿这个日志找奶牛“算账”的时候，发现这个日志被损坏了（有些部分被丢失了）。现在要帮助农夫判断：这个日志是否“合法”。如果“合法”，就输出最小可能和最大可能奶牛棚被破坏的天数；如果不“合法”，就输出-1。（当天日志被丢失的部分用-1表示）

 

题目分析：

这道题是一道水题，但是由于看错题目（英语渣的我），当场没有做出来。这道题大概是这样的：

1.第一天要么就是0，要么就是-1（被丢失）。如果不是0和-1，直接可以判断不合法。

2.我们可以根据日志上的其他剩余信息推算出其他天的日志信息，最简单的：第一天无论是0还是-1，一定是可以推算出是0。其他：比如：

这里第6天记录了一个“2”，代表前2天奶牛棚被坏了，所以第4天一定是0，第5天一定是1：

如果第4天不是0，或者第5天不是1，那么一定不合法：

我们再想一想：第7，8，9天我们能推出来吗？根据现有条件第7，8，9天的日志内容是不一定的：假如第6天后奶牛棚没有被破坏，那么后面的所有内容就可以被推出来，否则也不一定。所以现在就先要编写一个把能推的都推出来的代码，看究竟是否合法。

 

在写这个代码的时候，一般我们会从1遍历倒N，可是这里反过来遍历(N-1)代码会更容易些，为什么？假设第6天记录的是3，那么第5天是不是记录的是3-1，也就是2：

然后就一直进行减1的操作，直到是0为止：

这样想的结果是不是和刚刚按从1-N的顺序想的结果一样，但是代码只需要从(N-1)遍历一次就行了。

 

推算完全部结果后（这时肯定是合法的，如果中间检查到不合法就直接输出-1然后结束程序），奶牛棚被破坏的天数最少的天数就是推算出来后，日志内容是0的总天数。

（注：这里最少天数为2）

最多的天数怎样计算？只要把不能推算出来的日志内容全部变为0：

然后统计0的个数（不能推算出来日志内容的天数 + 最少天数）就是最大天数的（这里最大天数为6）。

 

 

AC代码：

 1 #include <cstdio>
 2 #include <iostream>
 3 using namespace std;
 4 int n, a[105];
 5
 6 int main(){
 7     cin >> n;
 8     for(int i = 1; i <= n; i++){
 9         cin >> a[i];
10     }
11
12     if(a[1] != 0 && a[1] != -1){
13         cout << -1 << endl;   //最简单的不合法情况
14         return 0;
15     }
16
17     a[1] = 0;    //这个不要漏
18     for(int i = n; i >= 1;){
19         while(a[i] == -1 && i >= 1) i--; //写这种代码时一定要记得 "i >= 1" 这样的限制条件
20         if(i == 0) break;  //遍历完
21
22         int t = a[i];
23         while(t >= 0 && i >= 1){
24             if(a[i] != t && a[i] != -1){  //推算出的不合法
25                 cout << -1 << endl;
26                 return 0;
27             }
28             a[i--] = t--;
29         }
30     }
31
32     int minn, cnt;
33     minn = cnt = 0;
34     for(int i = 1; i <= n; i++){  //简单的计数
35         if(a[i] == 0) minn++;
36         if(a[i] == -1) cnt++;
37     }
38
39     cout << minn << " " << minn+cnt << endl;
40     return 0;
41 }