제출 #153496

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
153496		brcode	Palindrome-Free Numbers (BOI13_numbers)	C++14	62.92 / 100	14 ms	10412 KiB

numbers

#include <iostream>
#include <vector>
#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
const long long MOD = 1e9+7;
vector<long long> v1;
long long dp[50][32][32][5][5];

long long rec(long long len,long long a,long long b,bool limit, int leading){
   // cout<<len<<" "<<a<<" "<<b<<" "<<limit<<" "<<leading<<endl;
    if(dp[len][a][b][limit][leading]!=-1){
        return dp[len][a][b][limit][leading];
    }
    if(len >= v1.size()-1){
        if(len == 3){

           
        }
        return 1;
    }
    if(limit){
        //last digit added was the largest possible digit
        dp[len][a][b][limit][leading] = 0;
        for(long long i=0;i<=9;i++){

            if(i>v1[len+1]){
                //can't add a larger digit
                break;
            }
            if(leading == 3 && i==0){
                //a new digit was just added after a bunch of zeroes
                if(i==v1[len+1]){
                    //largest possible digit I can add(Can add a 0, leading 0s don't count as a palindrome)
                    dp[len][a][b][1][leading] += rec(len+1,b,i,1,0);
                    dp[len][a][b][1][leading]%=MOD;

                }else{
                    //can do the same thing
                    dp[len][a][b][1][leading] += rec(len+1,b,i,0,0);
                    dp[len][a][b][1][leading]%=MOD;

                }
                continue;
            }
            else if(leading ==3){
                //a number that's not 0 is being added after a number that's just been added after a bunch of 0s
                if((i!=a && i!=b) && i == v1[len+1]){
                    //if this number's not a palindrome and it's the largest possible digit to add, add it
                    dp[len][a][b][1][leading] += rec(len+1,b,i,1,0);
                    dp[len][a][b][1][leading]%=MOD;
                    continue;
                }
                else if(i!=a && i!=b){
                    //same as above, but it's not the larges possible digit to add
                    dp[len][a][b][1][leading] += rec(len+1,b,i,0,0);
                    dp[len][a][b][1][leading]%=MOD;
                    continue;

                }
            }
            else if(leading && i==0){
                //have leading 0s, want to add another 0
                if(i==v1[len+1]){
                    //largest digit we can add is this 0(I don't think this case would actually exist)
                    dp[len][a][b][1][leading] += rec(len+1,b,i,1,1);
                    dp[len][a][b][1][leading]%=MOD;

                }else{
                    //we add this 0 and continue (our current string is just a bunch of 0s)
                    dp[len][a][b][1][leading] += rec(len+1,b,i,0,1);
                    dp[len][a][b][1][leading]%=MOD;

                }
                continue;
            }
            else if(leading){
                //we have a bunch of leading 0s and we want to add a non zero digit
                 if((i!=a && i!=b) && i == v1[len+1]){
                    //first 2 conditions will always hold true I think,the third is a check to see if this digit is the max digit we can add. If it is we shift our leading state to 3(dealt with above)
                    dp[len][a][b][1][leading] += rec(len+1,b,i,1,3);
                    dp[len][a][b][1][leading]%=MOD;
                    continue;
                }
                else if(i!=a && i!=b){
                    //Again first 2 conditions will always hold true
                    dp[len][a][b][1][leading] += rec(len+1,b,i,0,3);
                    dp[len][a][b][1][leading]%=MOD;
                    continue;

                }
            }
            else if((i!=a && i!=b) && i == v1[len+1]){
                //after a bunch of leading 0s(or none), we have a non zero digit + x more digits (where x+1>=2)
                dp[len][a][b][1][leading] += rec(len+1,b,i,1,0);
                dp[len][a][b][1][leading]%=MOD;

            }
            else if(i!=a && i!=b){
                dp[len][a][b][1][leading] += rec(len+1,b,i,0,0);
                dp[len][a][b][1][leading]%=MOD;

            }

        }
        dp[len][a][b][limit][leading]%=MOD;

        return dp[len][a][b][limit][leading];
    }else{
        dp[len][a][b][limit][leading] = 0;

        for(long long i=0;i<=9;i++){

            if(leading == 3 && i == 0){
                dp[len][a][b][limit][leading] += rec(len+1,b,i,limit,0);
                dp[len][a][b][limit][leading]%=MOD;
            }else if(leading ==3){
               
                if(i!=a && i!=b){

                    dp[len][a][b][limit][leading] += rec(len+1,b,i,limit,0);
                    dp[len][a][b][limit][leading]%=MOD;
                }
            }
            else if(leading && i==0){
                dp[len][a][b][limit][leading] += rec(len+1,b,i,limit,1);

                dp[len][a][b][limit][leading]%=MOD;
                continue;
            }else if(leading){
                dp[len][a][b][limit][leading] += rec(len+1,b,i,limit,3);

                dp[len][a][b][limit][leading]%=MOD;
                continue;
            }
            else if(i!=a && i!=b){
                dp[len][a][b][limit][leading] += rec(len+1,b,i,limit,0);
                dp[len][a][b][limit][leading]%=MOD;
            }

        }
        dp[len][a][b][limit][leading]%=MOD;
        return dp[len][a][b][limit][leading];
    }
}
long long solve(long long curr){
    v1.clear();
    for(long long i=0;i<=49;i++){
        for(long long j=0;j<=31;j++){
            for(long long k=0;k<=31;k++){
                for(long long l=0;l<=4;l++){
                    for(long long m=0;m<=4;m++){
                        dp[i][j][k][l][m] = -1;
                    }
                }
            }
        }
    }
    while(curr){
       v1.push_back(curr%10);

       curr/=10;
    }

    v1.push_back(0);
    reverse(v1.begin(),v1.end());
    long long ans = 0;
    for(long long i=0;i<=9;i++){
        if(v1[1]<i){
            break;
        }

        if(i ==0){
            ans+=rec(1,30,0,false,true);
            ans%=MOD;
           // cout<<ans<<endl;
            continue;
        }

        ans+=rec(1,30,i,v1[1] == i,false);
        ans%=MOD;
        //cout<<ans<<endl;

    }
    return ans;
}
int main(){
     long long a,b;
    cin>>a>>b;
    cout<<solve(b)-solve(a-1)<<endl;

}

컴파일 시 표준 에러 (stderr) 메시지

numbers.cpp: In function 'long long int rec(long long int, long long int, long long int, bool, int)':
numbers.cpp:15:12: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
     if(len >= v1.size()-1){
        ~~~~^~~~~~~~~~~~~~

• Subtask 1: 21.25 / 25
• Subtask 2: 41.666666666666664 / 75