B.
Motarack’s Birthday
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
inputCopy
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
outputCopy
1 11
5 35
3 6
0 42
0 0
1 2
3 4
題意：在n個數中，把數值爲-1的數 都填成數字x，使得k最小 k爲相鄰兩項之間 差的絕對值
可以這樣考慮，如果當前位置的數不是-1並且前一位是-1或者後一位是-1，我們就把當前這個數丟進數組s裏，m的取值就是排序後 s[0]+s[s.size()-1]/2,爲什麼這樣呢？
因爲和-1相鄰的數就要和m相減，所以我們就去算sb數組中最大值+最小值的和除以2 保證差值儘可能小
#include<bits/stdc++.h>
using namespace std;
int a[100005];
vector<int>s;
int main()
{
int t;
cin>>t;
while(t--)
{
s.clear();
int n;cin>>n;
for(int i=1;i<=n;i++) cin>>a[i];
for(int i=1;i<=n;i++){
if(a[i]!=-1 && (a[i-1]==-1 || a[i+1]==-1)) s.push_back(a[i]);
}
sort(s.begin(),s.end());
int m;
if(s.empty()) m=0;
else m=(s[0]+s[s.size()-1])/2;
int k=0;
for(int i=1;i<=n;i++){
if(a[i]==-1) a[i]=m;
if(i>=2) k=max(k,abs(a[i]-a[i-1]));
}
cout<<k<<" "<<m<<endl;
}
return 0;
}
C.
Ayoub’s function
Ayoub thinks that he is a very smart person, so he created a function f(s), where s is a binary string (a string which contains only symbols “0” and “1”). The function f(s) is equal to the number of substrings in the string s that contains at least one symbol, that is equal to “1”.
More formally, f(s) is equal to the number of pairs of integers (l,r), such that 1≤l≤r≤|s| (where |s| is equal to the length of string s), such that at least one of the symbols sl,sl+1,…,sr is equal to “1”.
For example, if s=“01010” then f(s)=12, because there are 12 such pairs (l,r): (1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,4),(3,5),(4,4),(4,5).
Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers n and m and asked him this problem. For all binary strings s of length n which contains exactly m symbols equal to “1”, find the maximum value of f(s).
Mahmoud couldn’t solve the problem so he asked you for help. Can you help him?
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤105) — the number of test cases. The description of the test cases follows.
The only line for each test case contains two integers n, m (1≤n≤109, 0≤m≤n) — the length of the string and the number of symbols equal to “1” in it.
Output
For every test case print one integer number — the maximum value of f(s) over all strings s of length n, which has exactly m symbols, equal to “1”.
Example
inputCopy
5
3 1
3 2
3 3
4 0
5 2
outputCopy
4
5
6
0
12
Note
In the first test case, there exists only 3 strings of length 3, which has exactly 1 symbol, equal to “1”. These strings are: s1=“100”, s2=“010”, s3=“001”. The values of f for them are: f(s1)=3,f(s2)=4,f(s3)=3, so the maximum value is 4 and the answer is 4.
In the second test case, the string s with the maximum value is “101”.
In the third test case, the string s with the maximum value is “111”.
In the fourth test case, the only string s of length 4, which has exactly 0 symbols, equal to “1” is “0000” and the value of f for that string is 0, so the answer is 0.
In the fifth test case, the string s with the maximum value is “01010” and it is described as an example in the problem statement.
簡單來說就是劃分0，1將0等分成k個區間，然後要用到容斥的思想我們先求出最大可能值，只需要減去不可能的情況就可
#include<bits/stdc++.h>
using namespace std;
int T;
long long n,m,len,num1,num2,ans;
int main(){
cin>>T;
while(T–){
cin>>n>>m;
len=(n-m)/(m+1);
num1=(n-m-len*(m+1));
num2=m+1-num1;
ans=n*(n-1)-len*(len+1)num1-len(len-1)*num2;
ans/=2;
cout<<ans+m<<endl;
}
}