time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Mishka is trying really hard to avoid being kicked out of the university. In particular, he was doing absolutely nothing for the whole semester, miraculously passed some exams so that just one is left.
There were nn classes of that subject during the semester and on ii-th class professor mentioned some non-negative integer aiai to the students. It turned out, the exam was to tell the whole sequence back to the professor.
Sounds easy enough for those who attended every class, doesn't it?
Obviously Mishka didn't attend any classes. However, professor left some clues on the values of aa to help out students like Mishka:
- aa was sorted in non-decreasing order (a1≤a2≤⋯≤ana1≤a2≤⋯≤an);
- nn was even;
- the following sequence bb, consisting of n2n2 elements, was formed and given out to students: bi=ai+an−i+1bi=ai+an−i+1.
Professor also mentioned that any sequence aa, which produces sequence bb with the presented technique, will be acceptable.
Help Mishka to pass that last exam. Restore any sorted sequence aa of non-negative integers, which produces sequence bb with the presented technique. It is guaranteed that there exists at least one correct sequence aa, which produces the given sequence bb.
Input
The first line contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the length of sequence aa. nn is always even.
The second line contains n2n2 integers b1,b2,…,bn2b1,b2,…,bn2 (0≤bi≤10180≤bi≤1018) — sequence bb, where bi=ai+an−i+1bi=ai+an−i+1.
It is guaranteed that there exists at least one correct sequence aa, which produces the given sequence bb.
Output
Print nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤10180≤ai≤1018) in a single line.
a1≤a2≤⋯≤ana1≤a2≤⋯≤an should be satisfied.
bi=ai+an−i+1bi=ai+an−i+1 should be satisfied for all valid ii.
Examples
input
Copy
4 5 6
output
Copy
2 3 3 3
input
Copy
6 2 1 2
output
Copy
0 0 1 1 1 2
题解：
题目大意是说b[i]=a[i]+a[n-i]的值，并且要保证是递增的，我们可以想，让左边的尽可能小，并且比右边的小，我们就可以去他左边的和右边较小的最大值即可，需要注意的是数据范围是 long long int型的
代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
int n;
cin>>n;
long long int k;
long long int a[200005];
for(int t=0;t<n/2;t++)
{
cin>>k;
if(t==0)
{
a[0]=0;
a[n-1]=k;
}
else
{
a[t]=max(a[t-1],min(a[n-t],k-a[n-t]));
a[n-1-t]=k-a[t];
}
}
for(int t=0;t<n;t++)
{
cout<<a[t]<<" ";
}
return 0;
}