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
分析：主席树套树状数组；
主席树可以看成一棵棵前缀权值线段树，单点修改p，就要改p及其之后所有线段树；
运用树状数组思想只需要修改log个即可；
注意内存严格，所以一开始静态建树，动态修改即可；
时间空间复杂度O(Nlog2N)；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <bitset> #include <map> #include <queue> #include <stack> #include <vector> #define rep(i,m,n) for(i=m;i<=n;i++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define sys system("pause") const int maxn=5e4+10; using namespace std; inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} inline void umax(ll &p,ll q){if(p<q)p=q;} inline void umin(ll &p,ll q){if(p>q)p=q;} inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m,k,t,a[maxn],b[60001],root[maxn],rx[maxn],lp[21],rp[21],ls[maxn*51],rs[maxn*51],s[maxn*51],sz,num; int q[10001],ql[10001],qr[10001],qk[10001]; void insert(int l,int r,int x,int &y,int z,int p) { y=++sz; s[y]=s[x]+p; if(l==r)return; int mid=l+r>>1; ls[y]=ls[x],rs[y]=rs[x]; if(z<=mid)insert(l,mid,ls[x],ls[y],z,p); else insert(mid+1,r,rs[x],rs[y],z,p); } void init(int x,int y) { lp[0]=rp[0]=0; while(x){lp[++lp[0]]=rx[x],x-=x&(-x);} while(y){rp[++rp[0]]=rx[y],y-=y&(-y);} } void add(int pos,int x,int cnt) { while(pos<=n) { insert(1,num,rx[pos],rx[pos],x,cnt); pos+=pos&(-pos); } } int gao(int l,int r,int x,int y,int z) { if(l==r)return l; int mid=l+r>>1; int p=0,q=0,i; rep(i,1,lp[0])p-=s[ls[lp[i]]]; rep(i,1,rp[0])p+=s[ls[rp[i]]]; q=s[ls[y]]-s[ls[x]]; if(z<=q+p) { rep(i,1,lp[0])lp[i]=ls[lp[i]]; rep(i,1,rp[0])rp[i]=ls[rp[i]]; return gao(l,mid,ls[x],ls[y],z); } else { rep(i,1,lp[0])lp[i]=rs[lp[i]]; rep(i,1,rp[0])rp[i]=rs[rp[i]]; return gao(mid+1,r,rs[x],rs[y],z-q-p); } } int main() { int i,j; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); num=0; sz=0; memset(rx,0,sizeof(rx)); rep(i,1,n)scanf("%d",&a[i]),b[++num]=a[i]; rep(i,1,m) { char str[2]; scanf("%s%d%d",str,&ql[i],&qr[i]); q[i]=str[0]=='Q'; if(q[i])scanf("%d",&qk[i]); else b[++num]=qr[i]; } sort(b+1,b+num+1); num=unique(b+1,b+num+1)-b-1; rep(i,1,n)a[i]=lower_bound(b+1,b+num+1,a[i])-b; rep(i,1,n)insert(1,num,root[i-1],root[i],a[i],1); rep(i,1,m) { if(q[i]) { init(ql[i]-1,qr[i]); printf("%d ",b[gao(1,num,root[ql[i]-1],root[qr[i]],qk[i])]); } else { qr[i]=lower_bound(b+1,b+num+1,qr[i])-b; add(ql[i],a[ql[i]],-1); add(ql[i],qr[i],1); a[ql[i]]=qr[i]; } } } return 0; }