이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
typedef long long ll;
#define FOR(i,x,y) for(ll i=x; i<y; i++)
#define FORNEG(i,x,y) for(ll i=x; i>y; i--)
#define double long double
using namespace std;
ll n,a,b;
ll nums[2005];
vector<ll> dp1[2005];
ll dp2[2005][2005];
ll poww(ll n){
ll sus = 1;
FOR(i,0,n){
sus *= 2;
}
return sus;
}
int check(ll v){
FOR(i,0,n+2) FOR(j,0,n+2) dp2[i][j] = 1000000000000000;
FOR(i,0,n+1) dp1[i] = {1000000000000000,1000000000000000};
dp1[0] = {0,0};
if (a==1){
FOR(i,0,n+1){
ll cur = 0;
FOR(x, i+1, n+1){
cur += nums[x];
if ((cur&v)==cur){
dp1[x] = {cur, min(dp1[x][1], dp1[i][1] + 1)};
}
}
}
if ((dp1[n][0]&v)==dp1[n][0] && dp1[n][1] <= b) return 1;
return 0;
}else{
dp2[0][0] = 0;
FOR(i,0,n+1){
FOR(j,0,n+1){
ll cur = 0;
FOR(x, i+1, n+1){
cur += nums[x];
if ((cur&v)==cur) dp2[x][j+1] = min(dp2[x][j+1], dp2[i][j] | cur);
}
}
}
FOR(i,a,b+1){
if ((dp2[n][i]&v) == dp2[n][i]) return 1;
}
}
return 0;
}
int main(){
cin >> n >> a >> b;
FOR(i,0,n) cin >> nums[i+1];
ll sus = poww(50)-1;
FORNEG(i, 49, -1){
sus -= poww(i);
if (check(sus)) continue;
else sus += poww(i);
}
cout << sus;
}
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