Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này

Code mẫu của flashmt

const maxl=16;
var l:longint; a,b:int64;
    f:array[0..maxl,0..9,1..maxl] of int64;
    s:array[0..maxl,0..9] of int64;
    sum,e:array[0..maxl] of int64;
    p,pp:array[-1..maxl] of int64;
    d:array[1..maxl] of longint;

procedure init;
var i,j:longint;
begin
     p[-1]:=0; p[0]:=1; p[1]:=9;
     pp[-1]:=0; pp[0]:=1; pp[1]:=10;
     for i:=2 to maxl do
     begin
          p[i]:=p[i-1]*10;
          pp[i]:=pp[i-1]*10;
     end;
end;

procedure dp;
var i,j,k:longint;
begin
     for j:=0 to 9 do
     begin
          f[1,j,1]:=j;
          s[1,j]:=j;
          sum[1]:=sum[1]+j;
     end;
     for i:=2 to maxl do
     begin
          for j:=0 to 9 do
          begin
               f[i,j,1]:=sum[i-1]-s[i-1,j]+p[i-1]*j;
               s[i,j]:=s[i,j]+f[i,j,1];
          end;
          for j:=0 to 9 do
              for k:=2 to i do
              begin
                   f[i,j,k]:=f[i-1,j,k-1]+j*(k shl 1-1)*p[i-k];
                   s[i,j]:=s[i,j]+f[i,j,k];
              end;
          for j:=0 to 9 do sum[i]:=sum[i]+s[i,j];
     end;
end;

procedure dem(a:int64;x:longint);
var i:longint; t,u,tt:int64;
begin
     e[0]:=a+1; t:=0; tt:=0;
     u:=pp[l-x+1]*d[x-1];
     for i:=1 to l-x+1 do
     begin
          u:=u div 10;
          t:=t+u; tt:=t+pp[l-x-i+1];
          if a<t then e[i]:=0
          else
              if a>=tt then e[i]:=pp[l-x-i+1]
              else e[i]:=a-t+1;
          e[i-1]:=e[i-1]-e[i];
     end;
end;

function calc(a:int64):int64;
var i,j,k:longint; re:int64; st:string;
begin
     re:=0; calc:=0;
     if a=0 then exit;
     str(a,st); l:=length(st);
     for i:=1 to l do d[i]:=ord(st[i])-48;
     for j:=0 to d[1]-1 do
         re:=re+s[l,j];
     a:=a mod pp[l-1];
     for i:=2 to l do
     begin
          fillchar(e,sizeof(e),0);
          dem(a,i);
          for k:=0 to l-i+1 do
              re:=re+d[i-1]*(k shl 1+1)*e[k];
          for j:=0 to d[i]-1 do
              re:=re+s[l-i+1,j];
          a:=a mod pp[l-i];
     end;
     re:=re+d[l];
     calc:=re;
end;

begin
     init;
     dp;
     read(a,b);
     writeln(calc(b)-calc(a-1));
end.

Code mẫu của ladpro98

#include <bits/stdc++.h>

using namespace std;

vector<int> D;
long long power[19];
long long memo[19][10][19][2];
long long value[19];

long long dp(int i, int digit, int cnt, bool smaller) {
    if (i < 0) return digit * cnt * cnt;
    long long &ans = memo[i][digit][cnt][smaller];
    if (ans != -1) return ans;
    ans = 0;
    for (int now = 0, limit = smaller ? 9 : D[i]; now <= limit; ++now) {
        if (now != digit) {
            long long mul;
            if (smaller || (now < D[i]))
                mul = power[i];
            else
                mul = i ? (value[i - 1] + 1) : 1;
            ans += dp(i - 1, now, 1, smaller || (now < D[i])) + mul * digit * cnt * cnt;
        } else {
            ans += dp(i - 1, now, cnt + 1, smaller || (now < D[i]));
        }
    }
    return ans;
}

long long solve(long long x) {
    if (x == 0) return 0;
    D.clear();
    while (x > 0) {
        D.push_back(x % 10);
        x /= 10;
    }
    value[0] = D[0];
    for (int i = 1; i < D.size(); ++i)
        value[i] = power[i] * D[i] + value[i - 1];
    memset(memo, -1, sizeof(memo));
    return dp(D.size() - 1, 0, 0, 0);
}


int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    power[0] = 1;
    for (int i = 1; i < 19; ++i) power[i] = power[i - 1] * 10;
    long long A, B;
    cin >> A >> B;
    cout << solve(B) - solve(A - 1) << endl;
    return 0;
}

Code mẫu của RR

#include <iostream>
#include <algorithm>

#define FOR(i,a,b) for(long i=a; i<=b; i++)
#define DOW(i,a,b) for(long i=a; i>=b; i--)
#define ll long long
using namespace std;

ll a,b,f[20][10][20][2],sl[20][10][20][2];
long cs[20];

void ptich(ll b) {
    memset(cs,0,sizeof cs);
    DOW(i,16,1) {
        cs[i]=b%10;
        b/=10;
    }
}

void update(long i,long last,long len,long ok1,long next,long ok2) {
    if (!sl[i][last][len][ok1]) return ;
    ll tmp=0LL;
    long len2;
    if (next==last) len2=len+1;
    else {
        len2=1;
        tmp=last*len*len;
    }

    f[i+1][next][len2][ok2]+=f[i][last][len][ok1]+tmp*sl[i][last][len][ok1];
    sl[i+1][next][len2][ok2]+=sl[i][last][len][ok1];
}

ll get(ll b) {
    if (!b) return 0LL;
    memset(sl,0,sizeof sl);
    memset(f,0,sizeof f);

    b=b*10+1;
    ptich(b);

    FOR(now,0,cs[1]-1) {
        f[1][now][1][1]=0LL;
        sl[1][now][1][1]=1LL;
    }
    f[1][cs[1]][1][0]=0LL;
    sl[1][cs[1]][1][0]=1LL;

    FOR(i,1,15)
    FOR(last,0,9)
    FOR(len,1,i) {
        FOR(next,0,9)
            update(i,last,len,1,next,1);
        FOR(next,0,cs[i+1]-1)
            update(i,last,len,0,next,1);
        update(i,last,len,0,cs[i+1],0);
    }

    ll res=0;
    FOR(len,1,16) res+=f[16][0][len][1];
    return res;
}

int main() {
//    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
    cin >>a >>b;
    ll tmp=0;
    if (b==1000000000000000LL) {
        tmp=1LL;
        b-=1LL;
    }
    cout<<get(b)-get(a-1);
    return 0;
}

Code mẫu của hieult

#include <cstdio>
#include <cstring>
//#include <iostream>
//#include <conio.h>
#include <algorithm>

using namespace std;

typedef long long ll;

ll A,B,mu10[17],f[17][12];

ll tim(int n,ll mu)
{
     if(n<0) return 0;
     ll u = max(mu*mu10[n],A);
     ll v = min((mu+1)*mu10[n]-1,B);
     if(u>v) return 0;
     return v-u+1;
}

ll xuly(int n, int so, ll mu)
{
     ll minn = mu*mu10[n];ll maxx = (mu+1)*mu10[n]-1;
     if(minn > B || maxx <A) return 0;
     if(minn >= A && maxx <= B && f[n][so] != -1) return f[n][so];
     ll kq = 0;
     for(int i = 0;i<=9;i++)
     {
          if(i== so) continue;
          ll t = mu;
          for(int k = 1;k<=n;k++)
          {
               t = t*10+i;
               kq = kq+i*k*k*(tim(n-k,t)-tim(n-k-1,t*10+i))+xuly(n-k,i,t);
          }
     }
     if(minn >= A && maxx <= B) f[n][so] = kq;
     return kq;
}

int main()
{

    scanf("%lld %lld",&A,&B);
    mu10[0] = 1;
    for(int i = 1;i<=15;i++) mu10[i] = mu10[i-1]*10;
    memset(f,-1,sizeof(f));
    printf("%lld\n",xuly(15,10,0));
    //getch();
}

Code mẫu của ll931110

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <fstream>
#include <iostream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <vector>
typedef long long ll;
using namespace std;

ll mem[16][11];
ll p10[16];
ll A,B;

ll calc(int n,ll prefix)
{
   if (n < 0) return 0;
   ll low = max(p10[n] * prefix,A);
   ll high = min(p10[n] * (prefix + 1) - 1,B);
   return (low > high) ? 0 : (high - low + 1);
};

ll rec(int n,int prev,ll prefix)
{
   ll low = p10[n] * prefix;
   ll high = p10[n] * (prefix + 1) - 1;
   if (high < A || B < low) return 0;
   if (A <= low && high <= B && mem[n][prev] != -1) return mem[n][prev];
   ll ret = 0;
   for (int digit = 0; digit < 10; digit++) if (digit != prev)
   {
       ll t = prefix;
       for (int k = 1; k <= n; k++)
       {
           t = t * 10 + digit;
           ret += digit * k * k * (calc(n - k,t) - calc(n - k - 1,t * 10 + digit)) + rec(n - k,digit,t);
       };
   };   
   if (A <= low && high <= B) mem[n][prev] = ret;  return ret;
};

int main()
{
//    freopen("tres.in","r",stdin);
//    freopen("tres.ou","w",stdout);
    cin >> A >> B;
    p10[0] = 1;
    for (int i = 1; i <= 15; i++) p10[i] = p10[i - 1] * 10;
    memset(mem,-1,sizeof(mem));
    cout << rec(15,10,0) << endl;
};

Code mẫu của khuc_tuan

#include <iostream>
using namespace std;

int a[22], na;
long long F[18][2][18][18];
long long pow10[18], value[18];

long long go(int pos, int isless, int last, int len) {
    if(pos==-1) return last * len * len;
    long long &ret = F[pos][isless][last][len];
    if(ret!=-1) return ret;
    ret = 0;
    for(int ad=0;ad<10;++ad)
        if(isless || ad <= a[pos])
            ret += go( pos - 1, isless || ad < a[pos], ad, (ad==last) ? (len + 1) : 1) + ((ad==last) ? 0 : (last * len * len)) * ((isless || ad < a[pos]) ? pow10[pos] : value[pos]);
    return ret;
}

long long calc(long long X) {
    if(X==0) return 0;
    na = 0;
    for(;X!=0;X/=10) a[na++] = X % 10;
    value[0] = 1;
    value[1] = a[0];
    for(long long i=1, hs = 10;i<na;++i, hs *= 10) value[i+1] = value[i] + a[i] * hs;
    for(int i=1;i<=na;++i) value[i] ++ ;
    memset( F, -1, sizeof(F));
    return go( na - 1, 0, 0, 0);
} 

int main() {
    pow10[0] = 1;
    for(int i=1;i<18;++i) pow10[i] = 10 * pow10[i-1];
    long long A, B;
    cin >> A >> B;
    cout << calc(B) - calc(A-1) << endl;
    return 0;
}