Question meaning: n number of m queries (n, m≤105n, m≤105)
Each query has l, r, k to ask how many numbers are less than or equal to k in the interval [l, r]
If you use the chairman tree to do it, you can query the i-th smallest number and compare it with k.
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx)
#include<bits/stdc++.h>
using namespace std;
#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
//#define ll long long
#define mp make_pair
#define pb push_back
#define sz size
#define fi first
#define se second
#define pf printf
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e2+7;
const int _=1e5+10;
const double EPS=1e-8;
const double eps=1e-8;
const LL mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}
inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}
void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//cerr << "run time is " << clock() << endl;
int T[_],L[_<<5],R[_<<5],sum[_<<5];
int tot=0;
void build(int &rt,int l,int r) {
rt=++tot;
sum[rt]=0;
if(l>=r) return ;
int mid=(l+r)>>1;
build(L[rt],l,mid);
build(R[rt],mid+1,r);
}
void update(int &rt,int l,int r,int last,int pos,int val) {
rt=++tot;
L[rt]=L[last];
R[rt]=R[last];
sum[rt]=sum[last]+val;
if(l>=r) return ;
int mid=(l+r)>>1;
if(pos<=mid) update(L[rt],l,mid,L[last],pos,val);
else update(R[rt],mid+1,r,R[last],pos,val);
}
int query(int u,int v,int l,int r,int k) {
if(l>=r) return l;
int mid=(l+r)>>1;
int tmp=sum[L[v]]-sum[L[u]];
if(tmp>=k)
return query(L[u],L[v],l,mid,k);
else
return query(R[u],R[v],mid+1,r,k-tmp);
}
int n,m;
int a[_],Hash[_];
void solve() {
s_2(n,m);
tot=0;
FOR(1,n,i) {
s_1(a[i]);
Hash[i]=a[i];
}
sort(Hash+1,Hash+1+n);
int d=unique(Hash+1,Hash+n+1)-Hash-1;
build(T[0],1,d);
FOR(1,n,i) {
int x=lower_bound(Hash+1,Hash+d+1,a[i])-Hash;
update(T[i],1,d,T[i-1],x,1);
}
W(m--) {
int l,r,k;
s_3(l,r,k);
l++,r++;
int ll=1,rr=r-l+1,ans=0;
W(ll<=rr) {
int mm=(ll+rr)>>1;
int tmp=Hash[query(T[l-1],T[r],1,d,mm)];
if(tmp<=k) {
if(mm==r-l+1||Hash[query(T[l-1],T[r],1,d,mm+1)]>k) {
ans=mm;
break;
}
ll=mm+1;
}
else {
if(mm==1||Hash[query(T[l-1],T[r],1,d,mm-1)]<=k){
ans=mm-1;
break;
}
rr=mm-1;
}
}
print(ans);
}
}
int main() {
//freopen( "1.in" , "r" , stdin );
//freopen( "1.out" , "w" , stdout );
int t=1;
//init();
s_1(t);
for(int cas=1;cas<=t;cas++) {
printf("Case %d:\n",cas);
solve();
}
}