Submission #731762

#	Time	Username	Problem	Language	Result	Execution time	Memory
731762		n0sk1ll	Palindrome-Free Numbers (BOI13_numbers)	C++17	100 / 100	1 ms	332 KiB

numbers

#include <bits/stdc++.h>

#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);cerr.tie(0)
#define mp make_pair
#define xx first
#define yy second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define all(x) x.begin(),x.end()
#define ff(i,a,b) for (int i = a; i < b; i++)
#define fff(i,a,b) for (int i = a; i <= b; i++)
#define bff(i,a,b) for (int i = b-1; i >= a; i--)
#define bfff(i,a,b) for (int i = b; i >= a; i--)

using namespace std;
long double typedef ld;
unsigned int typedef ui;
long long int typedef li;
pair<int,int> typedef pii;
pair<li,li> typedef pli;
pair<ld,ld> typedef pld;
vector<vector<int>> typedef graph;
unsigned long long int typedef ull;
//const int mod = 998244353;
const int mod = 1000000007;







//Note to self: Check for overflow

li dp[2][10][10]; //da li moram da postujem pravila, dvocifreni zavrsetak
li pp[2][10][10]; //prethodni dp

vector<int> cifre;

li uradi(li n)
{
    if (n<=9) return n+1;

    li temp=n;
    cifre.clear();
    while (temp)
    {
        cifre.pb(temp%10);
        temp/=10;
    }
    reverse(all(cifre));

    ff(STA_,0,2) ff(i,0,10) ff(j,0,10) dp[STA_][i][j]=0,pp[STA_][i][j]=0;

    ff(i,1,10) ff(j,0,10) if (i!=j)
    {
        if (i<cifre[0]) dp[0][i][j]++;
        else if (i==cifre[0] && j<cifre[1]) dp[0][i][j]++;
        else if (i==cifre[0] && j==cifre[1]) dp[1][i][j]++;
    }

    ff(i,2,(int)cifre.size())
    {

        ff(STA_,0,2) ff(i,0,10) ff(j,0,10) pp[STA_][i][j]=dp[STA_][i][j],dp[STA_][i][j]=0;
        ff(ac,0,10) ff(bc,0,10) ff(tc,0,10) if (tc!=ac && tc!=bc) //prekjucerasnja cifra, jucerasnja cifra, danasnja cifra
        {
            dp[0][bc][tc]+=pp[0][ac][bc];
            if (tc<cifre[i]) dp[0][bc][tc]+=pp[1][ac][bc];
            if (tc==cifre[i]) dp[1][bc][tc]+=pp[1][ac][bc];
        }
    }

    li ret=0;
    ff(i,0,10) ff(j,0,10) ret+=dp[0][i][j];
    ff(i,0,10) ff(j,0,10) ret+=dp[1][i][j];

    return ret;
}

li solve(li n)
{
    if (n==-1) return 0ll;
    if (n==0) return 1;
    if (n==1'000'000'000'000'000'000) return solve(n-1);

    li kur=0;
    li stp=1;
    while (stp>0)
    {
        stp*=10;
        if (n<=stp-1) return kur+uradi(n);
        kur+=uradi(stp-1);
    }

    return -69; //ovo zaista ne bi trebalo da se izvrsi
}

int main()
{
    FAST;

    li a,b; cin>>a>>b;
    cout<<solve(b)-solve(a-1)<<"\n";
}

//Note to self: Check for overflow

• Subtask 1: 25 / 25
• Subtask 2: 75 / 75