题目大意：给出一个序列，单点修改，询问区间第k大。

思路：如果不带修改，那么划分树就可以解决，但是划分树是静态的树，不支持修改。带修改的主席舒其实就是外层fenwick套内层权值线段树，但是权值线段树必须动态开节点。然后修改的时候就像树状数组修改那样，每次修改logn个权值线段树。查询的时候也一样，返回logn个权值线段树统计的和。

最后为了求区间第k大，还需要二分答案。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 10010
#define MAX_RANGE 1000000000
using namespace std;

struct Complex{
	Complex *son[2];
	int num;
	Complex() {
		son[0] = son[1] = NULL;
		num = 0;
	}
}*fenwick[MAX];

int cnt,asks;
int src[MAX];

char c[10];

inline void Fix(int x,int c);
inline void _Fix(int x,int c);
inline int GetSum(int x,int k);

void Insert(Complex *&a,int l,int r,int x);
void Delete(Complex *&a,int l,int r,int x);
int Ask(Complex *a,int l,int r,int k);

int Ask(int x,int y,int k);

int main()
{
	cin >> cnt >> asks;
	for(int i = 1;i <= cnt; ++i) {
		scanf("%d",&src[i]);
		Fix(i,src[i]);
	}
	for(int x,y,z,i = 1;i <= asks; ++i) {
		scanf("%s",c);
		if(c[0] == 'Q') {
			scanf("%d%d%d",&x,&y,&z);
			printf("%d\n",Ask(x,y,z));
		}
		else {
			scanf("%d%d",&x,&y);
			_Fix(x,src[x]);
			Fix(x,src[x] = y);
		}
	}
	return 0;
}

inline void Fix(int x,int c)
{
	for(;x <= cnt;x += x&-x)
		Insert(fenwick[x],0,MAX_RANGE,c);
}

inline void _Fix(int x,int c)
{
	for(;x <= cnt;x += x&-x) 
		Delete(fenwick[x],0,MAX_RANGE,c);
}

inline int GetSum(int x,int k)
{
	int re = 0;
	for(;x;x -= x&-x)
		re += Ask(fenwick[x],0,MAX_RANGE,k);
	return re;
}

int Ask(int x,int y,int k)
{
	int l = 0,r = MAX_RANGE,ans = 0;
	while(l <= r) {
		int mid = (l + r) >> 1;
		int temp = GetSum(y,mid) - GetSum(x - 1,mid);
		if(temp >= k)
			r = mid - 1,ans = mid;
		else	l = mid + 1;
	}
	return ans;
}

void Insert(Complex *&a,int l,int r,int x)
{
	if(a == NULL)	a = new Complex();
	a->num++;
	if(l == r)	return ;
	int mid = (l + r) >> 1;
	if(x <= mid)	Insert(a->son[0],l,mid,x);
	else	Insert(a->son[1],mid + 1,r,x);
}

void Delete(Complex *&a,int l,int r,int x)
{
	if(!--a->num) {
		free(a);
		a = NULL;
		return ;
	}
	if(l == r)	return ;
	int mid = (l + r) >> 1;
	if(x <= mid)	Delete(a->son[0],l,mid,x);
	else	Delete(a->son[1],mid + 1,r,x);
}

int Ask(Complex *a,int l,int r,int k)
{
	if(a == NULL)	return 0;
	if(l == r)	return a->num;
	int mid = (l + r) >> 1;
	if(k <= mid)	return Ask(a->son[0],l,mid,k);
	else {
		int left = a->son[0] == NULL ? 0:a->son[0]->num;
		return left + Ask(a->son[1],mid + 1,r,k);
	}
}

BZOJ 1901 Zju 2112 Dynamic Rankings 带修改主席树