이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
using namespace std;
int rez[400005];
int krez;
const int INF = 1e8;
/**
n = 4
m = 11
1 1 1 -4 1 1 1 -4 1 2
*/
void solve(int n, int m)
{
if(n==m)
{
krez=n-1;
for(int i=1;i<n;i++)
rez[i]=1;
return;
}
if(m%n==0)
{
krez=m-1;
for(int i=1;i<m;i++)
rez[i]=-1;
return;
}
if(n==2)
{
krez=m;
for(int i=1;i<=m;i++)
{
if(i%2==1)
rez[i]=INF;
else
rez[i]=-(INF+1);
}
return;
}
while(m%n!=n-1)
m++;
krez=m;
int ult=n;
for(int i=n;i<=m;i+=n)
{
for(int j=i-n+1;j<i;j++)
rez[j]=1;
rez[i]=-n;
ult=i;
}
for(int j=ult+1;j<=m;j++)
rez[j]=1;
}
signed main()
{
ios_base::sync_with_stdio(0);cin.tie(0);
int t;
cin>>t;
while(t--)
{
int n,m;
cin>>n>>m;
bool bl=0;
if(n>m)
{
swap(n,m);
bl=1;
}
solve(n,m);
cout<<krez<<"\n";
for(int i=1;i<=krez;i++)
{
if(!bl)
cout<<rez[i]<<" ";
else
cout<<-rez[i]<<" ";
}
cout<<"\n";
}
return 0;
}
/**
pref[i] - pref[i-n] < 0
pref[i] - pref[i-m] > 0
pref[i-n] > pref[i]
pref[i-m] < pref[i]
n = 4
m = 11
1 1 1 -4 1 1 1 -4 1 1 1
m - 2 * m/n > 0
n = 2
m = 11
5 -6 5 -6 5 -6 5
*/
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