Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2) D. Restore Permutation(线段树)

 

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

An array of integers p1,p2,…,pn is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2],[1],[1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2],[1,1],[2,3,4].

There is a hidden permutation of length n.

For each index i, you are given si, which equals to the sum of all pj such that j

Your task is to restore the permutation.

Input

The first line contains a single integer n (1≤n≤2⋅105) — the size of the permutation.

The second line contains n integers s1,s2,…,sn (0≤si≤n(n−1)2).

It is guaranteed that the array s corresponds to a valid permutation of length n.

Output

Print n integers p1,p2,…,pn — the elements of the restored permutation. We can show that the answer is always unique.

Examples

input

Copy

3
0 0 0

output

Copy

3 2 1

input

Copy

2
0 1

output

Copy

1 2

input

Copy

5
0 1 1 1 10

output

Copy

1 4 3 2 5

Note

In the first example for each i there is no index j satisfying both conditions, hence si are always 0.

In the second example for i=2 it happens that j=1 satisfies the conditions, so s2=p1.

In the third example for i=2,3,4 only j=1 satisfies the conditions, so s2=s3=s4=1. For i=5 all j=1,2,3,4 are possible, so s5=p1+p2+p3+p4=10.

【思路】

其实这题是个简单线段树，不过根据cf的尿性我一般做不到这来，也就是没有时间看，卡C题，c题玄学题一大堆人过。

每次从最小的一个赋值，然后更新这个点的对后面提供的答案消去，同时消掉这个点

#include
#define rep(i,a,b) for(int i=a;i<=(b);++i)
#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int N=2e5+10;
const ll inf=1e11;
struct node
{
    ll mn;
    int i;
    bool operator <(const node&o)const
    {
        return mn<=o.mn;
    }
}a[4*N];
ll lazy[4*N];
int n;
void build(int id,int l,int r)
{
    if(l==r)
    {
        scanf("%lld",&a[id].mn);
        a[id].i=l;
        return ;
    }
    int mid=l+r>>1;
    build(id<<1,l,mid);
    build(id<<1|1,mid+1,r);
    a[id]=min(a[id<<1],a[id<<1|1]);
}
void pushdown(int id)
{
    if(lazy[id])
    {
        lazy[id<<1]+=lazy[id];
        lazy[id<<1|1]+=lazy[id];
        a[id<<1].mn+=lazy[id];
        a[id<<1|1].mn+=lazy[id];
        lazy[id]=0;
    }
}
void up(int id,int l,int r,int ql,int qr,ll val)
{
    if(ql<=l&&r<=qr)
    {
        lazy[id]+=val;
        a[id].mn+=val;
        return ;
    }
    pushdown(id);
    int mid=l+r>>1;
    if(ql<=mid) up(id<<1,l,mid,ql,qr,val);
    if(qr>mid) up(id<<1|1,mid+1,r,ql,qr,val);
    a[id]=min(a[id<<1],a[id<<1|1]);
}
int ans[N];
int main()
{
    cin>>n;
    build(1,1,n);
    rep(i,1,n)
    {
        ans[a[1].i]=i;
        int id=a[1].i;
        up(1,1,n,id,id,inf);
        up(1,1,n,id+1,n,-i);
    }
    rep(i,1,n) printf("%d ",ans[i]);
}