Codeforces Global Round 5 部分题解

tourist的劲题，掉了17分，着实心痛，原本是有分可恰到的

A、给你一个数列\(a\)，你须要构造一个数列\(b\)，使得每个\(b\)都等于\(a/2\)，向上或向下取整由你决定

而且还要使得\(\Sigma b\)的总和等于0

数据是保证有解的

那么咱们就令全部\(b\)等于\(\lfloor a/2 \rfloor\)，求出总和

再遍历一遍数组，sum过大则把某些负数调整为向上取整，不然把正数调整

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a) 
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define nmsl cout<<"NMSL"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
    char ch=getchar();int s=0,w=1;
    while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}
    while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
    return s*w;
}
inline void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);
}
int gcd(int a, int b){
    if(a==0) return b;
    if(b==0) return a;
    if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
    else if(!(b&1)) return gcd(a,b>>1);
    else if(!(a&1)) return gcd(a>>1,b);
    else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}

int n,a[maxn],b[maxn];

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>n;re(i,1,n) cin>>a[i];
    int sum=0;
    re(i,1,n) b[i]=a[i]/2,sum+=b[i];
    re(i,1,n){
        if(a[i]%2==0) continue;
        if(sum==0) break;
        if(sum<0){
            if(a[i]>0){
                sum-=b[i];
                sum+=(a[i]+1)/2;
                b[i]=(a[i]+1)/2;
            }
        }
        else if(sum>0){
            if(a[i]<0){
                sum-=b[i];
                sum+=(a[i]-1)/2;
                b[i]=(a[i]-1)/2;
            }
        }
    }
    re(i,1,n) cout<<b[i]<<endl;
//  cout<<"? "<<sum<<endl;
    return 0;
}
/*
2
-3 3
*/

B、给你一些车辆进入隧道和离开隧道的顺序，问哪些车超车了，即出隧道的位置比进隧道的位置更靠前

相似于二维偏序，维护一下每一个车的编号出现的位置便可

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a) 
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define nmsl cout<<"NMSL"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
    char ch=getchar();int s=0,w=1;
    while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}
    while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
    return s*w;
}
inline void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);
}
int gcd(int a, int b){
    if(a==0) return b;
    if(b==0) return a;
    if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
    else if(!(b&1)) return gcd(a,b>>1);
    else if(!(a&1)) return gcd(a>>1,b);
    else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}

int n,a[maxn],b[maxn];
int p[maxn],p2[maxn];

int tr[maxn];
void add(int x,int y=1){
    for(;x<=n;x+=lowbit(x)) tr[x]+=y;
}
int sum(int x){
    int res=0;
    for(;x>0;x-=lowbit(x)) res+=tr[x];
    return res;
}
bool v[maxn];

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>n;re(i,1,n) cin>>a[i],p[a[i]]=i;
    re(i,1,n) cin>>b[i],p2[b[i]]=i;
    re(i,1,n){
        add(p2[a[i]]);
        if(sum(n)-sum(p2[a[i]])>=1) v[a[i]]=1;
    }
    int ans=0;
    re(i,1,n) if(v[i]) ans++;
    cout<<ans;
    return 0;
}

C一、C二、给你偶数个点，你须要按某种顺序删除全部的点对

好比说你如今准备删除点\(i\)和点\(j\)，那么必须保证不存在点\(z\)，使得点\(z\)的三个坐标都位于\(i\)和\(j\)的三个坐标构成的闭区间内

考虑暴力，每次取出距离最近的一堆点便可，时间复杂度\(O(n^2)\)

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a) 
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define nmsl cout<<"NMSL"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
    char ch=getchar();int s=0,w=1;
    while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}
    while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
    return s*w;
}
inline void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);
}
int gcd(int a, int b){
    if(a==0) return b;
    if(b==0) return a;
    if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
    else if(!(b&1)) return gcd(a,b>>1);
    else if(!(a&1)) return gcd(a>>1,b);
    else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}

int n;
struct node{
    db x,y,z;
    int id;
}o[maxn];
vep v;
bool del[maxn];
vector<pair<db,pii>> tmp;

bool cmp(node a,node b){
    if(a.x!=b.x) return a.x<b.x;
    else{
        if(a.y!=b.y) return a.y<b.y;
        else return a.z<b.z;
    }
}

db L(node a,node b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>n;
    re(i,1,n) cin>>o[i].x>>o[i].y>>o[i].z,o[i].id=i;
    re(i,1,n){
        re(j,1,n){
            if(i==j) continue;
            tmp.pub(mkp(L(o[i],o[j]),mkp(o[i].id,o[j].id)));
        }
    }
    sort(tmp.begin(),tmp.end());
    fo(i,0,tmp.size()){
        if(del[tmp[i].snd.fst]||del[tmp[i].snd.snd]) continue;
        del[tmp[i].snd.fst]=1;
        del[tmp[i].snd.snd]=1;
        v.emplace_back(tmp[i].snd.fst,tmp[i].snd.snd);
    }
    for(auto i:v) cout<<i.fst<<' '<<i.snd<<endl;
    return 0;
}

考虑优化，首先把全部点按三个坐标排序

对于全部x和y坐标相等的点，从中间向两边取，而后按一样的方法取x坐标相同的点

最后剩余的点按顺序取便可

算法很简单，下标写错就被hack

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a) 
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define nmsl cout<<"NMSL"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
    char ch=getchar();int s=0,w=1;
    while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}
    while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
    return s*w;
}
inline void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);
}
int gcd(int a, int b){
    if(a==0) return b;
    if(b==0) return a;
    if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
    else if(!(b&1)) return gcd(a,b>>1);
    else if(!(a&1)) return gcd(a>>1,b);
    else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}

int n;
struct node{
    int x,y,z,id;
}o[maxn];
vep v;
vei tmp;
bool del[maxn];

bool cmp(node a,node b){
    if(a.x!=b.x) return a.x<b.x;
    else{
        if(a.y!=b.y) return a.y<b.y;
        else return a.z<b.z;
    }
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>n;
    re(i,1,n) cin>>o[i].x>>o[i].y>>o[i].z,o[i].id=i;
    sort(o+1,o+1+n,cmp);
    re(i,1,n){
        int l1=i;
        while(l1<n&&o[l1].x==o[l1+1].x&&o[l1].y==o[l1+1].y) l1++;
        if(l1-i+1>=2){
            int l=(i+l1)/2,r=l+1;
            while(l>=i&&r<=l1){
                v.pub(mkp(o[l].id,o[r].id));
                del[o[l].id]=del[o[r].id]=1;
                l--,r++;
            }
        }
        i=l1;
    }
    vei tt;
    re(i,1,n){
        int l2=i;
        tt.clear();
        while(l2<=n&&o[l2].x==o[i].x){
            if(!del[o[l2].id]) tt.pub(o[l2].id);
            l2++;
        }
        if(tt.size()>=2){
            int l=tt.size()/2-1,r=l+1;
            while(l>=0&&r<tt.size()){
                v.pub(mkp(tt[l],tt[r]));
                del[tt[l]]=del[tt[r]]=1;
                l--,r++;
            }
        }
        i=l2-1;
    }
    re(i,1,n){
        if(!del[o[i].id]) tmp.pub(o[i].id);
    }
    int l=tmp.size()/2-1,r=l+1;
    while(l>=0&&r<tmp.size()){
        v.pub(mkp(tmp[l],tmp[r]));
        l--,r++;
    }
    for(auto i:v) cout<<i.fst<<' '<<i.snd<<endl;
    return 0;
}
/*
4
32117556 7831842 -38931866
32117556 -42244261 -38931866
-24487444 -42244261 -38931866
-24487444 7831842 -38931866
*/

D、有n首歌构成的歌单，每首歌有一个开心值\(a[i]\)，第\(n\)首歌曲结束以后会当即播放第\(1\)首歌曲

若是已经播放的歌曲的最大开心值为\(x\)，若是下一首歌曲的开心值\(a[i]\)，知足$a[i]< \lceil x/2 \rceil $，那么你会当即中止播放

问你从每一个位置开始播放了几首歌曲以后停下，一直不停下输出-1

首先意识到：只要有一首歌不停下，那么全部的歌都不会停下

感性的说，若是有一首歌永远都不会停下，那么就说明它把全部的歌曲都播放了一遍

那么只须要从1遍历到\(3n\)便可，注意这里应该设为\(3n\)，即把数字复制三遍，不然样例二过不去

剩下的考虑答案的某些性质：

假设当前歌曲的播放区间为\([L,R]\)，那么下一首歌的播放中止位置不会比\(R\)更小，不然前一首歌必定会在那个更小的终点处停下

而后用ST表维护区间最大值就搞定了

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a) 
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define nmsl cout<<"NMSL"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
    char ch=getchar();int s=0,w=1;
    while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}
    while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
    return s*w;
}
inline void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);
}
int gcd(int a, int b){
    if(a==0) return b;
    if(b==0) return a;
    if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
    else if(!(b&1)) return gcd(a,b>>1);
    else if(!(a&1)) return gcd(a>>1,b);
    else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}

int n,a[maxn];
bool all=1;
int ans[maxn];
int dp[maxn][35];

void build(){
    re(i,1,3*n)
        dp[i][0]=a[i];
    re(j,1,30){
        for(int i=1;i+(1LL<<(j-1))<=3*n;++i)
            dp[i][j]=max(dp[i][j-1],dp[i+(1LL<<(j-1))][j-1]);
    }
}

int ask(int l,int r){
    if(l==r) return a[l];
    int k=log2(r-l+1);
    return max(dp[l][k],dp[r-(1LL<<k)+1][k]);
}

void gao(){
    build();
    int l=1,r=l+1;
    while(l<=n){
        if(r==l) r++;
//      cout<<l<<' '<<a[r]<<' '<<(ask(l,r-1)+1)/2<<endl;
        while(r<=3*n&&a[r]>=(ask(l,r-1)+1)/2) r++;
        ans[l]=r-l;
        l++;
    }
    re(i,1,n) cout<<ans[i]<<' ';
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>n;re(i,1,n) cin>>a[i],a[i+n]=a[i],a[i+2*n]=a[i];
    int x=(a[1]+1)/2;
    re(i,2,3*n){
        if(a[i]<x){
            all=0;
            break;
        }
        x=max(x,(a[i]+1)/2);
    }
    if(all){
        re(i,1,n) cout<<-1<<' ';
        return 0;
    }
    gao();
    return 0;
}