gym101239J Pipe Stream

题意

已知 \(x\) 满足 \(l\leq x\leq r\) ，现在每次能询问一个 \(y(0\leq y\leq s)\) ，第 \(i\) 次会回答 \(ix\geq y\)是否成立。

求最少的次数，使得能确定 \(x\) 在一个长度 \(\leq t\) 的区间内。

\(q\leq 100\)。

题解

考虑从 \(r\) 开始往下每长度 \(t\) 分一段，现在变成判断 \(x\) 在哪一段内。

考虑把询问看成在一个有段数个叶子的bst上搜索，其中要求第\(i\)个叶子的深度不超过\(d_i\)。

那么我们从上到下每次加一层，每次把自由叶子数 \(\times 2\)，然后把必须放这一层的放在这一层(\(-\)个数)，直到够放剩下的叶子即可。

正确性显然。

代码

/*
Author: QAQAutoMaton
Lang: C++
Code: G.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef complex<double> cp;
typedef pair<int,int> pii;
int inf;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T,class T2>int chkmin(T &a,T2 b){return a>b?a=b,1:0;}
template<class T,class T2>int chkmax(T &a,T2 b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T,class T2>T mmin(T a,T2 b){return a<b?a:b;}
template<class T,class T2>T mmax(T a,T2 b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
template<class T>bool sort2(T &a,T &b){return a>b?swap(a,b),1:0;}
#define min mmin
#define max mmax
#define abs aabs
struct __INIT__{
	__INIT__(){
		fill((unsigned char*)&inf,(unsigned char*)&inf+sizeof(inf),0x3f);
	}
}__INIT___;
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	template<typename A>
	inline bool read (A &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		while((x=gc())==' '||x=='\n' || x=='\r');
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	template<typename A>
	inline bool write (A x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
namespace run{
	int s,l,r,t,c;
	int xcount(int x){
		return (r-x)/t;
	}
	int get(int c){
		return max(xcount(max(s/(c+1)+1,l))-xcount(min(s/c+1,r)),0);
//		r/c .. r/(c+1)	
	}
	bool main(){
		read(s,l,r,t,c);
		l*=c;r*=c;t*=c;
		if(r-l<=t)return write("0\n");
		int w=(r-l-1)/t+1,cur=1,cnt=0;
		if(r-t>s)return write("impossible\n");
		while(cur<w){
			if(cur<0)return write("impossible\n");
			++cnt;	
			cur<<=1;
			int c=get(cnt);
			w-=c;
			cur-=c;
		}
		return write(cnt,'\n');
	}
}
signed main(){
#ifdef QAQAutoMaton 
	freopen("G.in","r",stdin);
	freopen("G.out","w",stdout);
#endif
	int t;
	read(t);
	for(;t;--t)run::main();
	return 0;
}