Misha walked through the snowy forest and he was so fascinated by the trees to decide to draw his own tree!
Misha would like to construct a rooted tree with n
vertices, indexed from 1 to n, where the root has index 1. Every other vertex has a parent pi, and i is called a child of vertex
.
Below there is a tree with 6
vertices. The subtree of vertex 2 contains vertices 2, 3, 4, 5. Hence the size of its subtree is 4
.
The branching coefficient of the tree is defined as the maximum number of children in any vertex. For example, for the tree above the branching coefficient equals
. Your task is to construct a tree with n vertices such that the sum of the subtree sizes for all vertices equals s
, and the branching coefficient is minimum possible.
InputThe only input line contains two integers n
and s — the number of vertices in the tree and the desired sum of the subtree sizes (2≤n≤105; 1≤s≤1010).
OutputIf the required tree does not exist, output «No». Otherwise output «Yes» on the first line, and in the next one output integers p2
, p3, ..., pn, where pi denotes the parent of vertex i.
Examples Input3 5Output
Yes 1 1Input
4 42Output
NoInput
6 15Output
Yes 1 2 3 1 5Note
Below one can find one of the possible solutions for the first sample case. The sum of subtree sizes equals 3+1+1=5
, and the branching coefficient equals 2.
Below one can find one of the possible solutions for the third sample case. The sum of subtree sizes equals 6+3+2+1+2+1=15
, and the branching coefficient equals 2.
題意：給定N，S，讓你構造一個大小為N的數，使得每個節點子樹大小之和為S，如果存在，請構造一個樹，使得兒子最多的點的兒子數量（P）最少。
思路：我們發現對於大小一定的樹，越瘦長K越大（一條鏈，最大為N*(N+1)/2）,越矮胖越小（菊花樹，最小為N+N-1），那麼如果K不在這個範圍我們輸出-1；如果在，我們一定看i有構造一個滿足題意的樹。 我們可以二分得到P。然後來構造。 我的構造方式是先構造一條鏈，此時的sum=N*(N+1)/2；如果sum>S，我們就把最下面的點移到上面的某個位置，知道sum=S。
#include<bits/stdc++.h> #define ll long long #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int maxn=1000010;ll N,S; ll fa[maxn],q[maxn],d[maxn],head,tail,sz[maxn],son[maxn]; bool check(ll Mid) { ll tN=N,now=1,p=1,res=0; while(tN){ res+=min(p,tN)*now; if(res>S) return false; tN-=min(p,tN); p*=Mid; now++; } return true; } int main() { cin>>N>>S; ll Mn=N+N-1; ll Mx=N*(N+1)/2; if(S<Mn||S>Mx) return puts("NO"),0; ll L=1,R=N-1,Mid,res; while(L<=R){ Mid=(L+R)/2; if(check(Mid)) res=Mid,R=Mid-1; else L=Mid+1; } puts("YES"); rep(i,1,N) sz[i]=1; ll Now=Mx,D=2; for(int i=N;;i--){ if(Now==S) break; if(sz[D]==sz[D-1]*res) D++; if(Now-S>=i-D){ sz[D]++; sz[i]--; Now-=(i-D); } else { sz[i]--; sz[i-(Now-S)]++; Now=S; } } head=tail=1; q[head]=1; d[1]=1; ll p=1; rep(i,2,N) { if(sz[i]==0) break; L=p+1; R=p+sz[i]; rep(j,L,R){ while(d[q[head]]!=i-1||son[q[head]]==res){ head++; } fa[j]=q[head]; son[q[head]]++; q[++tail]=j; d[j]=i; } p=R; } rep(i,2,N) printf("%lld ",fa[i]); return 0; }