이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
int k;
long long l, r;
long long dp[50][15][15][5][5][2];
vector<int> digitl;
vector<int> digitr;
long long solve (int cur, int last1, int last2, bool low, bool high, bool ok)
{
//if (last1>=0&&last2>=0)
if (dp[cur][last1][last2][low][high][ok]!=-1) return dp[cur][last1][last2][low][high][ok];
if (cur==k) return 1ll;
long long cnt=0ll;
for (int i=0; i<10; i++)
if (i!=last1&&i!=last2&&(low||i>=digitl[cur])&&(high||i<=digitr[cur]))
cnt+=solve(cur+1, (!ok&&i<1?10:i), last1, low||i>digitl[cur], high||i<digitr[cur], ok||(i>0));
dp[cur][last1][last2][low][high][ok]=cnt;
return cnt;
}
int main()
{
//freopen("numbers.inp", "r", stdin);
//freopen("numbers.out", "w", stdout);
cin>>l>>r;
while (l>0)
digitl.push_back(l%10ll), l/=10ll;
while (r>0)
digitr.push_back(r%10ll), r/=10ll;
while ((int)digitl.size()<(int)digitr.size()) digitl.push_back(0);
while ((int)digitr.size()<(int)digitl.size()) digitr.push_back(0);
k=(int)digitr.size();
reverse(digitl.begin(), digitl.end());
reverse(digitr.begin(), digitr.end());
memset(dp, -1, sizeof dp);
cout<<solve(0, 10, 10, 0, 0, 0);
return 0;
}
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