CodeForces - 1081E Missing Numbers （平方数，构造）

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假设我们已经构造出了前2i - 2项，目前构造第 2i - 1项，而2i项已知，我们新插入的2i-1项要与他们满足如下关系

#define int ll
int v[MX];
signed main()
{
    int n;cin>>n;
    n>>=1;
    memset(v,-1,sizeof(v));
    rpp(i,n) cin>>v[i*2];
    int now=0;
    rpp(i,n) 
    {
        int y=v[i*2];
        for(int _=1;_*_<=y;++_)//下划线派qwq
        {
            if(y%_==0&&((y/_+_)%2==0)&&((y/_-_)%2==0)) 
            {
                int a=(y/_-_)/2;
                if(a*a<=now) continue;
                if(v[i*2-1]==-1) v[i*2-1]=a*a-now;
                else v[i*2-1]=min(v[i*2-1],a*a-now);
            }
        }
        if(v[i*2-1]==-1)
        {
            cout<<"No\n";
            stop;
            return 0;
        }
        now+=v[i*2-1]+v[i*2];
    }
    n<<=1;
    cout<<"Yes\n";
    rpp(i,n) cout<<v[i]<<" ";
    cout<<endl; 
    stop;
    return 0;
}