C ++, n = 6
작은 최적화로 무차별 대입.
#include<cassert>
#include<iostream>
#include<vector>
// ===========
/** Helper struct to print binary representation.
`std::cout<<bin(str,len)` prints (str:len) == the bitstring
represented by last (len) bits of (str).
*/
struct bin{
int str,len;
bin(int str,int len):str(str),len(len){}
};
std::ostream& operator<<(std::ostream& str,bin a){
if(a.len)
return str<<bin(a.str>>1,a.len-1)<<char('0'+(a.str&1));
else if(a.str)
return str<<"...";
else
return str;
}
// ===========
/// A patten of (len) bits of ones.
int constexpr pat1(int len){
return (1<<len)-1;
}
// TODO benchmark: make (res) global variable?
/**Append all distinct (subseqs+(sfx:sfxlen)) of (str:len)
with length (sublen) to (res).
*/
void subseqs_(
int str,int len,int sublen,
int sfx,int sfxlen,
std::vector<int>& res
){
// std::cout<<"subseqs_ : str = "<<bin(str,len)<<", "
// "sublen = "<<sublen<<", sfx = "<<bin(sfx,sfxlen)<<'\n';
assert(len>=0);
if(sublen==0){ // todo remove some branches can improve perf?
res.push_back(sfx);
return;
}else if(sublen==len){
res.push_back(str<<sfxlen|sfx);
return;
}else if(sublen>len){
return;
}
if(str==0){
res.push_back(sfx);
return;
}
int nTrail0=0;
for(int ncut;str&&nTrail0<sublen;
++nTrail0,
ncut=__builtin_ctz(~str)+1, // cut away a bit'0' of str
// plus some '1' bits
str>>=ncut,
len-=ncut
){
ncut=__builtin_ctz(str)+1; // cut away a bit'1' of str
subseqs_(str>>ncut,len-ncut,sublen-nTrail0-1,
sfx|1<<(sfxlen+nTrail0),sfxlen+nTrail0+1,
res
); // (sublen+sfxlen) is const. TODO global var?
}
if(nTrail0+len>=sublen) // this cannot happen if len<0
res.push_back(sfx);
}
std::vector<int> subseqs(int str,int len,int sublen){
assert(sublen<=len);
std::vector<int> res;
if(__builtin_popcount(str)*2>len){ // too many '1's, flip [todo benchmark]
subseqs_(pat1(len)^str,len,sublen,0,0,res);
int const p1sublen=pat1(sublen);
for(int& r:res)r^=p1sublen;
}else{
subseqs_(str,len,sublen,0,0,res);
}
return res;
}
// ==========
/** Append all distinct (supersequences+(sfx:sfxlen)) of (str:len)
with length (suplen) to (res).
Define (a) to be a "supersequence" of (b) iff (b) is a subsequence of (a).
*/
void supseqs_(
int str,int len,int suplen,
int sfx,int sfxlen,
std::vector<int>& res
){
assert(suplen>=len);
if(suplen==0){
res.push_back(sfx);
return;
}else if(suplen==len){
res.push_back(str<<sfxlen|sfx);
return;
}
int nTrail0; // of (str)
if(str==0){
res.push_back(sfx);
// it's possible that the supersequence is '0000..00'
nTrail0=len;
}else{
// str != 0 -> str contains a '1' bit ->
// supersequence cannot be '0000..00'
nTrail0=__builtin_ctz(str);
}
// todo try `nTrail0=__builtin_ctz(str|1<<len)`, eliminates a branch
// and conditional statement
for(int nsupTrail0=0;nsupTrail0<nTrail0;++nsupTrail0){
// (nsupTrail0+1) last bits of supersequence matches with
// nsupTrail0 last bits of str.
supseqs_(str>>nsupTrail0,len-nsupTrail0,suplen-1-nsupTrail0,
sfx|1<<(nsupTrail0+sfxlen),sfxlen+nsupTrail0+1,
res);
}
int const strMatch=str?nTrail0+1:len;
// either '1000..00' or (in case str is '0000..00') the whole (str)
for(int nsupTrail0=suplen+strMatch-len;nsupTrail0-->nTrail0;){
// because (len-strMatch)<=(suplen-1-nsupTrail0),
// (nsupTrail0<suplen+strMatch-len).
// (nsupTrail0+1) last bits of supersequence matches with
// (strMatch) last bits of str.
supseqs_(str>>strMatch,len-strMatch,suplen-1-nsupTrail0,
sfx|1<<(nsupTrail0+sfxlen),sfxlen+nsupTrail0+1,
res);
}
// todo try pulling constants out of loops
}
// ==========
int n,b;
std::vector<char> done;
unsigned min_undone=0;
int result;
void backtrack(int nchoice){
assert(!done[min_undone]);
++nchoice;
std::vector<int> supers_s;
for(int s:subseqs(min_undone,n,n-b)){
// obviously (s) is not chosen. Try choosing (s)
supers_s.clear();
supseqs_(s,n-b,n,0,0,supers_s);
for(unsigned i=0;i<supers_s.size();){
int& x=supers_s[i];
if(!done[x]){
done[x]=true;
++i;
}else{
x=supers_s.back();
supers_s.pop_back();
}
}
unsigned old_min_undone=min_undone;
while(true){
if(min_undone==done.size()){
// found !!!!
result=std::min(result,nchoice);
goto label1;
}
if(not done[min_undone])
break;
++min_undone;
}
if(nchoice==result){
// backtrack more will only give worse result
goto label1;
}
// note that nchoice is already incremented
backtrack(nchoice);
label1: // undoes the effect of (above)
for(int x:supers_s)
done[x]=false;
min_undone=old_min_undone;
}
}
int main(){
std::cin>>n>>b;
done.resize(1<<n,0);
result=1<<(n-b); // the actual result must be less than that
backtrack(0);
std::cout<<result<<'\n';
}
로컬로 실행하십시오.
[user202729@archlinux golf]$ g++ -std=c++17 -O2 delbits.cpp -o delbits
[user202729@archlinux golf]$ time for i in $(seq 1 3); do ./delbits <<< "6 $i"; done
12
4
2
real 0m0.567s
user 0m0.562s
sys 0m0.003s
[user202729@archlinux golf]$ time ./delbits <<< '7 1'
^C
real 4m7.928s
user 4m7.388s
sys 0m0.173s
[user202729@archlinux golf]$ time for i in $(seq 2 3); do ./delbits <<< "7 $i"; done
6
2
real 0m0.040s
user 0m0.031s
sys 0m0.009s
b > 0추가 입력 요구 사항을 가정 하고 있습니까? 또는 것n=3및b=0간단하게 출력2^n결과?