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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <random>
using namespace std;
using namespace __gnu_pbds;
#pragma GCC optimize("O3,Ofast,unroll-loops")
//#pragma GCC target("popcnt")
//#define int long long
#define ll long long
#define ld long double
#define endl '\n'
#define pii pair<int,int>
#define debug(...) printf(__VA_ARGS__)
//#define debug(...)
#define trace(x) cout<<#x<<"="<<x<<endl;
//#define trace(x)
#define vi vector<int>
#define popcnt(x) (int)__builtin_popcount(x)
#define vpii vector<pii >
#define vvi vector<vector<int> >
#define umap unordered_map
#define uset unordered_set
#define mk make_pair
#define pb push_back
#define pob pop_back
#define _ <<' '<<
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define read(x) for (auto& zu: (x)) cin >> zu;
#define write(x) for (auto& zu: (x))cout << zu<<' '; cout<<endl;
static inline void print(const int &x) {
if (x > 9)print(x / 10);
putchar('0' + x % 10);
}
#define F first
#define S second
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define mustawa ios::sync_with_stdio(false);cin.tie(nullptr);
#define mishari main
const int inf=1e9+7;
void fa(bool ok){ cout << (ok ? "YES": "NO") << endl;}
int pw(ll a, ll b, ll mod= 1e9 + 7){
ll result = 1;
while(b) {
if(b&1) {
result = (result*a)%mod;
}
a = (a * a) % mod;
b>>=1;
}
return result;
}
inline void usaco(string s) {
freopen((s + ".in").c_str(), "r", stdin);
freopen((s + ".out").c_str(), "w", stdout);
}int gcd(int a,int b){
if (b == 0)return a;
return gcd(b, a % b);
}
int lcm(int a, int b){
return (a / gcd(a, b)) * b;
}
//long double log(int base,int x){
// return log2(x)/log(base);
//}
int log(int base,int x){
return (int)(log2(x)/log(base));
}
void vod(){}
int sum(int l,int r){
if(l>r)swap(l,r);
int a=((l-1)*(l))/2;
int b=(r*(r+1))/2;
return b-a;
}
int L[(int)2e5+5],R[(int)(2e5+5)],dp[(int)2e5+5],n;
int r0(int i){
if(i<0)return 0;
if(dp[i]!=-1)return dp[i];
return dp[i]=max(r0(i-1),r0(i-L[i]-1));
}
void solve(){
cin>>n;
bool b=0;
for(int i=0;i<n;i++){
cin>>L[i]>>R[i];
if(L[i]!=R[i])b=1;
if(i&&L[i]!=L[i-1])b=1;
}if(n<=1000){
for(int i=0;i<n;i++){
dp[i]=1;
for(int j=0;j<i-L[i];j++){
if(j+R[i]<i)dp[i]=max(dp[i],dp[j]+1);
}
}return cout<<dp[n-1]<<endl,vod();
}
if(!b)return cout<<(n-1)/(L[0]+1)+1<<endl,vod();
else {
memset(dp,-1,sizeof dp);
cout<<r0(n-1)<<endl;
}
}
signed mishari() {
mustawa;
//usaco("nocross");
int tescases = 1;
//cin>>tescases;
for (int i = 1; i <= tescases; i++) {
//cout<<"Case "<<i<<":"<<endl;
solve();
}
}
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