B. Find The Array
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an array [a1,a2,…,an] such that 1≤ai≤109. Let S be the sum of all elements of the array a.
Let’s call an array b of n integers beautiful if:
1≤bi≤109 for each i from 1 to n;
for every pair of adjacent integers from the array (bi,bi+1), either bi divides bi+1, or bi+1 divides bi (or both);
2∑i=1n|ai−bi|≤S.
Your task is to find any beautiful array. It can be shown that at least one beautiful array always exists.
Input
The first line contains one integer t (1≤t≤1000) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2≤n≤50).
The second line contains n integers a1,a2,…,an (1≤ai≤109).
Output
For each test case, print the beautiful array b1,b2,…,bn (1≤bi≤109) on a separate line. It can be shown that at least one beautiful array exists under these circumstances. If there are multiple answers, print any of them.
Example
inputCopy
4
5
1 2 3 4 5
2
4 6
2
1 1000000000
6
3 4 8 1 2 3
outputCopy
3 3 3 3 3
3 6
1 1000000000
4 4 8 1 3 3
写的很莫名其妙，不知道怎么说，就是前后两项要满足相除乘本身要小于1，否则就把它变成1
#include<iostream>
#include<string>
#include<map>
#include<algorithm>
#include<memory.h>
#include<cmath>
#include<assert.h>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
int a[maxn],b[maxn];
vector<int>l,r;
void solve(){
ll t;
cin>>t;
while(t--){
ll n;
cin>>n;
ll sum=0;
for(int i=1;i<=n;i++){
cin>>a[i];
sum+=a[i];
}
b[1]=a[1];
for(int i=2;i<=n;i++)
{
if(a[i]%b[i-1]==0){
b[i]=a[i];
}
else
{
if(fabs(a[i]-a[i]/b[i-1]*b[i-1])<fabs(a[i]-1))
b[i]=a[i]/b[i-1]*b[i-1];
else{
b[i]=1;
}
}
}
for(int i=1;i<=n;i++){
cout<<b[i]<<' ';
}
cout<<endl;
}
}
int main()
{
ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
solve();
return 0;
}