이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#define fast ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#include <bits/stdc++.h>
using namespace std;
#define sqr 200
#define mid (l+r)/2
#define pb push_back
#define ppb pop_back
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define ins insert
#define era erase
#define C continue
#define mem(dp,i) memset(dp,i,sizeof(dp))
#define mset multiset
typedef long long ll;
typedef short int si;
typedef long double ld;
typedef pair<int,int> pi;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;
const ll mod=1e9+7;
const ll inf= 4e18;
const ld pai=acos(-1);
#include "molecules.h"
int n ,L,R;
vi v;
int a[100009];
bool dp[10009][10009];
bool done[10009][10009];
bool bt(int id,int sum){
if ( id == n ){
return (L<=sum && sum<=R);
}
bool &ret=dp[id][sum];
if(done[id][sum])return ret;
ret = 0;
ret = max ( bt(id+1,sum) , bt(id+1,sum+a[id]) );
return ret;
}
void path(int id,int sum){
if ( id == n)return ;
bool ans = bt(id,sum);
bool ret1=bt(id+1,sum);
if ( ret1 == ans){
path(id+1,sum);
return ;
}
v.pb(id);
path(id+1,sum+a[id]);
}
vi find_subset(int l, int u, vi w) {
n=w.size();
for(int i =0 ;i < n;i ++ ){
a[i]=w[i];
}
L =l ,R = u;
if ( bt(0,0) ){
path(0,0);
return v;
}
return std::vector<int>(0);
}
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