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>
using namespace std;
//------------------------------DEFINE------------------------------
//******************************************************************
#define IOS ios_base::sync_with_stdio(false); cin.tie(0),cout.tie(0)
#define ll long long
#define pb push_back
#define F first
#define S second
#define INF 1e18
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define pii pair<int,int>
#define pll pair<ll,ll>
#define OK cout<<"Ok"<<endl;
#define MOD (ll)(1e9+7)
#define endl "\n"
//******************************************************************
//----------------------------FUNCTION------------------------------
//******************************************************************
ll gcd(ll a,ll b){
if(a>b) swap(a,b);
if(a==0) return a+b;
return gcd(b%a,a);
}
ll lcm(ll a,ll b){
return a/gcd(a,b)*b;
}
bool is_prime(ll n){
ll k=sqrt(n);
if(n==2) return true;
if(n<2||n%2==0||k*k==n) return false;
for(int i=3;i<=k;i+=2){
if(n%i==0){
return false;
}
}
return true;
}
//*****************************************************************
//--------------------------MAIN-CODE------------------------------
const int mxn=1e6+5;
ll t=1,n,k,p[mxn];
ll funk(ll n,ll k){
if(n==0){
if(k%2==0){
return 2;
}
else{
return 1;
}
}
if(k%2==0){
return (p[n]+funk(n-1,k/2))%MOD;
}
else{
return funk(n-1,(k+1)/2)%MOD;
}
}
void solve(){
cin>>n>>k;
p[0]=1;
for(int i=1;i<=n;i++){
p[i]=(p[i-1]*2)%MOD;
}
cout<<funk(n-1,k)<<endl;
}
int main(){
IOS;
//cin>>t;
while(t--){
solve();
}
return 0;
}
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