R2_A_Taming the Herd

最新推荐文章于 2024-05-26 22:53:15 发布

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在本篇博客中，我们将探讨如何解决一个有趣的算法问题：修复受损的奶牛出逃记录，通过分析剩余的记录来确定奶牛可能的最少和最多出逃次数。这个问题涉及到逻辑推理和序列分析，我们将分享一种有效的解决方案。

题面

Taming the Herd

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 00 that day; if the most recent breakout was 33 days ago, the counter would read 33. 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 NN (1≤N≤1001≤N≤100), denoting the number of days since Farmer John started logging the cow breakout counter.

The second line contains NN space-separated integers. The iith integer is either −1−1, indicating that the log entry for day ii is missing, or a non-negative integer aiai (at most 100100), indicating that on day ii the counter was at aiai.

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 11, output a single integer −1−1. Otherwise, output two space-separated integers mm followed by MM, where mm is the minimum number of breakouts of any consistent sequence of events, and MM 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.

题目大意

J用数字记录下了奶牛什么时候进行破坏。比如1是代表1天前进行了破坏，0代表今天进行破坏。但是这个记录损坏了，出现了一些-1，-1可以代表任意数字。任务是求奶牛进行破坏的次数的最大值与最小值。如果这份记录有矛盾，则输出-1。

题目分析

根据那些大于0的数字，我们可以去确定一些-1的具体数值。那么我们可以由那些数字倒着推回来。每当我们得到一个大于0的数n，那么前n天的记录也就清楚了，唯一确定了。当前n天的中如果出现有0的出现，那么我们比如可以确定，这份记录有矛盾。

代码

#include <cstdio>
using namespace std;
const int maxn = 1e5;
typedef long long ll;
int arr[107];
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  for(int i = 0; i < n; i++)
    scanf("%d", &arr[i]);
  arr[0] = 0;
  bool flag = false;
  for(int i = 0; i < n; i++){
    // printf("%d\n", i);
    if(flag) break;

    if(arr[i] > 0){
      if(i - arr[i] < 0){flag = true; break;}
      if(arr[i - arr[i]] > 0) {
        flag = true;
        break;
      }else{
        for(int j = i - arr[i], cnt = 0; j < i; j++, cnt++){
          if(arr[j] == 0 && j != i - arr[i]){flag = true; break;}
          arr[j] = cnt;
        }
      }
    }
  }
  if(flag)
    printf("%d\n", -1);
  else{
    int fu = 0, zero = 0;
    for(int i = 0; i < n; i++){
      if(arr[i] == 0)
        zero++;
      if(arr[i] == -1)
        fu++;
    }
    printf("%d %d\n", zero, zero + fu);
  }
  return 0;
}