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.
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.
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
Sample Output
3
6
3
6
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 50010;
const int M = 10010;
const int BC = 233;
struct Block
{
int l, r;
};
Block B[M];
int arr[N], belong[N], unit, bcnt, b[N];
int n, m;
void reset(int x)
{
for (int i = B[x].l; i <= B[x].r; ++i)
b[i] = arr[i];
sort(b + B[x].l, b + B[x].r + 1);
}
void init()
{
unit = sqrt(n);
bcnt = n / unit;
if (n % unit)
++bcnt;
int i;
for (i = 1; i <= bcnt; ++i)
{
B[i].l = (i - 1) * unit + 1;
B[i].r = i * unit;
}
B[bcnt].r = n;
for (i = 1; i <= n; ++i)
belong[i] = (i - 1) / unit + 1;
for (i = 1; i <= bcnt; ++i)
reset(i);
}
void update(int x, int t)
{
int bx = belong[x];
arr[x] = t;
reset(bx);
}
int bs(int x, int key)
{
int l = B[x].l, r = B[x].r;
int ans = -1;
while (l <= r)
{
int mid = MID(l, r);
if (b[mid] <= key)
{
ans = mid;
l = mid + 1;
}
else
r = mid - 1;
}
return ~ans ? ans - B[x].l + 1 : 0;
}
int query(int l, int r, int k)
{
int L = 0, R = 1e9;
int ans = 1;
int bl = belong[l], br = belong[r], i;
while (L <= R)
{
int tk = 0;
int mid = MID(L, R);
for (i = l; i <= B[bl].r; ++i)
if (arr[i] <= mid)
++tk;
for (i = B[br].l; i <= r; ++i)
if (arr[i] <= mid)
++tk;
for (i = bl + 1; i < br; ++i)
tk += bs(i, mid);
if (tk >= k)
{
ans = mid;
R = mid - 1;
}
else
L = mid + 1;
}
return ans;
}
int main(void)
{
int tcase, i;
char ops[3];
int l, r, k, x, t;
scanf("%d", &tcase);
while (tcase--)
{
scanf("%d%d", &n, &m);
for (i = 1; i <= n; ++i)
scanf("%d", &arr[i]);
init();
for (i = 1; i <= m; ++i)
{
scanf("%s", ops);
if (ops[0] == 'Q')
{
scanf("%d%d%d", &l, &r, &k);
printf("%d
", query(l, r, k));
}
else
{
scanf("%d%d", &x, &t);
update(x, t);
}
}
}
return 0;
}