E. K Integers 題解(線段樹+思維)
阿新 • • 發佈:2021-11-10
題目連結
題目思路
假設\(1-k\)都挨在一起,那麼答案不就是逆序對嘛
但是要先把他們挨在一起
怎麼挨在一起呢,顯然是要放在中位數最合適
權值線段樹上二分就可以確定中位數了
然後再亂求求就行了
程式碼
不擺爛了,寫題#include<bits/stdc++.h> #define fi first #define se second #define debug cout<<"I AM HERE"<<endl; using namespace std; typedef long long ll; typedef unsigned long long ull; #define pii pair<long long,long long > //typedef pair<long long,long long > pii const int maxn=2e5+5,inf=0x3f3f3f3f,mod=1e9+7; const double eps=1e-6; const ll INF=0x3f3f3f3f3f3f3f3f; int n; int a[maxn],pos[maxn]; ll ni=0; ll tree[maxn<<2][2]; void push(int node){ tree[node][0]=tree[node<<1][0]+tree[node<<1|1][0]; tree[node][1]=tree[node<<1][1]+tree[node<<1|1][1]; } void update(int node,int pos,int l,int r){ if(l==r){ tree[node][0]++; tree[node][1]=l; return ; } int mid=(l+r)/2; if(mid>=pos) update(node<<1,pos,l,mid); else update(node<<1|1,pos,mid+1,r); push(node); } int serch(int node,int l,int r,int need){ if(l==r){ return l; } int mid=(l+r)/2; if(tree[node<<1][0]>=need) return serch(node<<1,l,mid,need); else return serch(node<<1|1,mid+1,r,need-tree[node<<1][0]); } ll query(int node,int L,int R,int l,int r,int op){ if(L>R) return 0; if(L<=l&&r<=R){ return tree[node][op]; } int mid=(l+r)/2; ll sum=0; if(mid>=L) sum+=query(node<<1,L,R,l,mid,op); if(mid<R) sum+=query(node<<1|1,L,R,mid+1,r,op); return sum; } signed main(){ scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d",&a[i]); pos[a[i]]=i; } for(int i=1;i<=n;i++){ ni+=query(1,pos[i]+1,n,1,n,0); update(1,pos[i],1,n); ll mid=serch(1,1,n,(i+1)/2); ll num1=query(1,1,mid-1,1,n,0); ll num2=query(1,mid+1,n,1,n,0); ll sum1=query(1,1,mid-1,1,n,1); ll sum2=query(1,mid+1,n,1,n,1); ll tmp1=(mid-1-num1+1 + mid-1)*num1/2-sum1; ll tmp2=sum2-(mid+1 + mid+1+num2-1)*num2/2; printf("%lld ",ni+tmp1+tmp2); } return 0; }