云华100的专栏

裸的划分树，也可以用其他方法。#include#include#define N 100100using namespace std;struct node{ int x,y,mid;}a[N*4];struct Tree{ int val,num,p;}tree[20][N];int st[N];void build(int x,int y,int ce

#include#include#define N 100100using namespace std;struct node{ int x,y,mid;}a[N*4];struct Tree{ int val,num,p;}tree[20][N];int st[N];void build(int x,int y,int cen,int t){ in

Minimum SumTime Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2721    Accepted Submission(s): 627Problem DescriptionYou are given

#include#include#include#define N 100100using namespace std;int ans,val[20][N],lsum[20][N],num[20][N],sum[N],st[N]; //num数组记录元素所在区间的当前位置之前进入左孩子的个数 //lsum数组记录元素所在区间进入左孩子的元素的和void build(int x,in

/*划分树+二分，二分第k个数题意：给一列数..若干个询问..问(l,r,h)...在[l,r]范围内..有多少个数小于等于h*/#include#include#include#define N 100100using namespace std;int st[N],val[20][N],num[20][N];void build(int x,int y,int cen){