Discription
Dark is going to attend Motarack’s birthday. Dark decided that the gift he is going to give to Motarack is an array a of n non-negative integers.
Dark created that array 1000 years ago, so some elements in that array disappeared. Dark knows that Motarack hates to see an array that has two adjacent elements with a high absolute difference between them. He doesn’t have much time so he wants to choose an integer k (0≤k≤109) and replaces all missing elements in the array a with k.
Let m be the maximum absolute difference between all adjacent elements (i.e. the maximum value of |ai−ai+1| for all 1≤i≤n−1) in the array a after Dark replaces all missing elements with k.
Dark should choose an integer k so that m is minimized. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of the test cases follows.
The first line of each test case contains one integer n (2≤n≤105) — the size of the array a.
The second line of each test case contains n integers a1,a2,…,an (−1≤ai≤109). If ai=−1, then the i-th integer is missing. It is guaranteed that at least one integer is missing in every test case.
It is guaranteed, that the sum of n for all test cases does not exceed 4⋅105.
Output
Print the answers for each test case in the following format:
You should print two integers, the minimum possible value of m and an integer k (0≤k≤109) that makes the maximum absolute difference between adjacent elements in the array a equal to m.
Make sure that after replacing all the missing elements with k, the maximum absolute difference between adjacent elements becomes m.
If there is more than one possible k, you can print any of them.
Example
input
7
5
-1 10 -1 12 -1
5
-1 40 35 -1 35
6
-1 -1 9 -1 3 -1
2
-1 -1
2
0 -1
4
1 -1 3 -1
7
1 -1 7 5 2 -1 5
output
1 11
5 35
3 6
0 42
0 0
1 2
3 4
Note
In the first test case after replacing all missing elements with 11 the array becomes [11,10,11,12,11]. The absolute difference between any adjacent elements is 1. It is impossible to choose a value of k, such that the absolute difference between any adjacent element will be ≤0. So, the answer is 1.
In the third test case after replacing all missing elements with 6 the array becomes [6,6,9,6,3,6].
|a1−a2|=|6−6|=0;
|a2−a3|=|6−9|=3;
|a3−a4|=|9−6|=3;
|a4−a5|=|6−3|=3;
|a5−a6|=|3−6|=3.
So, the maximum difference between any adjacent elements is 3.
題意
有一串非負數，有一些數因為某種原因丢失。要求在丢失處填入一個相同的數，讓所有的數中的兩個相鄰的數之間的內插補點最小。
輸出最大的內插補點和填入的數。
思路
周遊數組，找-1兩邊內插補點最大一組，取填入的數為兩邊的平均數。
在不是-1的某數左邊或右邊不等于-1時，記錄內插補點，取最大值。
最後內插補點與插入的數的左右的數的內插補點比較，取最大值。
AC代碼
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define pd(n) printf("%d\n", (n))
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define rep(i, a, n) for(int i=a; i<=n; i++)
#define per(i, n, a) for(int i=n; i>=a; i--)
int T;
int n;
int a[100001];
int c,ans;
int main()
{
scanf("%d",&T);
while(T--)
{
sd(n);
for(int i=1; i<=n; i++)
sd(a[i]);
a[0]=-1,a[n+1]=-1,c=0;
int k=1,tmp=-1,tt=0,q=1,f=-1;
int l=1e9+1;
int r=0;
for(int i=1;i<=n;i++)
{
if(a[i]!=-1)
{
f=a[i];
if(a[i+1]!=-1)
c=max(abs(a[i+1]-a[i]),c);
if(a[i-1]!=-1)
c=max(abs(a[i]-a[i-1]),c);
//continue;
}
else
{
if(a[i-1]!=-1)
{
l=min(a[i-1],l);
r=max(a[i-1],r);
}
if(a[i+1]!=-1)
{
l=min(a[i+1],l);
r=max(a[i+1],r);
}
}
}
ans=(l+r+1)/2;
c=max(abs(r-ans),c);
c=max(abs(ans-l),c);
if(f==-1)
{
cout<<0<<" "<<42<<endl;
continue;
}
cout<<c<<" "<<ans<<endl;
}
return 0;
}