The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.
Your task is to write a program for this computer, which
- Reads N numbers from the input (1 <= N <= 50,000)
- Processes M instructions of the input (1 <= M <= 10,000). These instructions include querying the k-th smallest number of a[i], a[i+1], ..., a[j] and change some a[i] to t.
Input
The first line of the input is a single number X (0 < X <= 4), the number of the test cases of the input. Then X blocks each represent a single test case.
The first line of each block contains two integers N and M, representing N numbers and M instruction. It is followed by N lines. The (i+1)-th line represents the number a[i]. Then M lines that is in the following format
Q i j k or
C i t
It represents to query the k-th number of a[i], a[i+1], ..., a[j] and change some a[i] to t, respectively. It is guaranteed that at any time of the operation. Any number a[i] is a non-negative integer that is less than 1,000,000,000.
There're NO breakline between two continuous test cases.
<b< dd="">
Output
For each querying operation, output one integer to represent the result. (i.e. the k-th smallest number of a[i], a[i+1],..., a[j])
There're NO breakline between two continuous test cases.
<b< dd="">
Sample Input
2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
题解：
树状数组的每个节点都是一颗线段树，但这棵线段树不再保存每个前缀的信息了，而是由树状数组的sum函数计算出这个前缀的信息，那么显而易见这棵线段树保存的是辅助数组S的值，即S=A[i-lowbit+1]+...+A[i],其中A[i]表示值为i的元素出现的次数。
那么对于每次修改，我们要修改树状数组上的logn棵树，对于每棵树，我们要修改logn个结点，所以时空复杂度为
O(（n+q）*logn*logn)，由于这道题n比较大，查询次数q比较小，所以我们可以初始时建立一颗静态的主席树，树状数组只保存每次修改的信息，那么时空复杂度降为了O（n*logn+q*logn*logn）
参考代码：
#include<bits/stdc++.h> using namespace std; #define lowbit(x) (x&-x) typedef long long ll; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return x*f; } const int maxn=6e4+10; struct Tree{ int ls,rs; int sum; } node[maxn*34]; struct Qnode{ bool flag; int l,r,k; } qn[maxn]; int T,n,q,a[maxn],b[maxn],num,rt[maxn]; char str[2]; int ul[maxn],ur[maxn],ca[maxn],s[maxn],cnt; void Build(int rt,int l,int r) { rt=++cnt; node[rt].sum=0; if(l==r) return ; int mid=l+r>>1; Build(node[rt].ls,l,mid); Build(node[rt].rs,mid+1,r); } void Update(int y,int &x,int l,int r,int pos,int val) { node[++cnt]=node[y];node[cnt].sum+=val;x=cnt; if(l==r) return ; int mid=l+r>>1; if(pos<=mid) Update(node[y].ls,node[x].ls,l,mid,pos,val); else Update(node[y].rs,node[x].rs,mid+1,r,pos,val); } void Add(int x,int val) { int res=lower_bound(b+1,b+1+num,a[x])-b; while(x<=n) { Update(s[x],s[x],1,num,res,val); x+=lowbit(x); } } int Sum(int x,bool temp) { int res=0; while(x>0) { if(temp) res+=node[node[ur[x]].ls].sum; else res+=node[node[ul[x]].ls].sum; x-=lowbit(x); } return res; } int Query(int L,int R,int x,int y,int l,int r,int k) { if(l==r) return l; int mid=l+r>>1; int res=Sum(R,true)-Sum(L,false)+node[node[y].ls].sum-node[node[x].ls].sum; if(k<=res) { for(int i=R;i;i-=lowbit(i)) ur[i]=node[ur[i]].ls; for(int i=L;i;i-=lowbit(i)) ul[i]=node[ul[i]].ls; return Query(L,R,node[x].ls,node[y].ls,l,mid,k); } else { for(int i=R;i;i-=lowbit(i)) ur[i]=node[ur[i]].rs; for(int i=L;i;i-=lowbit(i)) ul[i]=node[ul[i]].rs; return Query(L,R,node[x].rs,node[y].rs,mid+1,r,k-res); } } int main() { T=read(); while(T--) { n=read();q=read(); cnt=num=0; memset(rt,0,sizeof(rt)); for(int i=1;i<=n;++i) a[i]=read(),b[++num]=a[i]; for(int i=1;i<=q;++i) { scanf("%s",str); if(str[0]=='Q') { qn[i].flag=true; qn[i].l=read();qn[i].r=read();qn[i].k=read(); } else { qn[i].flag=false; qn[i].l=read();qn[i].r=read();b[++num]=qn[i].r; } } sort(b+1,b+1+num); int tot=unique(b+1,b+1+num)-b-1; num=tot; for(int i=1;i<=n;++i) ca[i]=lower_bound(b+1,b+1+num,a[i])-b; Build(rt[0],1,num); for(int i=1;i<=n;++i) Update(rt[i-1],rt[i],1,num,ca[i],1); for(int i=1;i<=n;++i) s[i]=rt[0]; for(int i=1;i<=q;++i) { if(qn[i].flag) { for(int j=qn[i].r;j;j-=lowbit(j)) ur[j]=s[j]; for(int j=qn[i].l-1;j;j-=lowbit(j)) ul[j]=s[j]; printf("%d ",b[Query(qn[i].l-1,qn[i].r,rt[qn[i].l-1],rt[qn[i].r],1,num,qn[i].k)]); } else { Add(qn[i].l,-1); a[qn[i].l]=qn[i].r; Add(qn[i].l,1); } } } return 0; }