Two Pointers Technique & Binary Search For Beginners

PROBLEM:
   Given n digits number-N. (n<=12, so that ans always fit in 64-bit int) .
   We have to calculate the sum of all of its permutation numbers.
for example:
N= 123. Then, n is 3 (digits.)
123
132
213
231
312
321
------
1332

Here, if we fixed '1' in first postion,then we can arrange remaining digits in (n-1)! ways.
so, 1 can be in pos-0 in (n-1)! times.      so, add- (1* fac[n-1]*dig[pos] )=1*2*1
      1 can be in pos-1 in (n-1)! times.       so,add = 1*2*10=20;
      1 can be in pos-2 in (n-1)! times.        so,add = 1*2*100=200;
Overall for 1=220.
similarly, for 2=444
for 3=666
--------------
so, ans= 1322.

NOW, consider if digits are not unique then ??
----->> Nothing to be amazed, Just multiply -- (n-1)! by occurace of remaining digits except fixed one.

Suppose for 3 digit number - 1 1 2

Here, if we fix 1 then,  for every digit position multiply ans & add- 1*2!*dig[pos...]. = 220
 finally, if we fix 2 then, for every digit pos multiply ans & add- 2* 2!/2!(repetation of 1) *dig[pos.] =220
 So, the answer is 444.
 
Suppose for 10 digits number as below:
1 4 4 1 2 2 6 6 6 1
after fixing -'1' ==> 9!/2!(remaining two 1s)*2!(for 4)*2!(for 2)*3!(for 6) .
similarly, fix -2,4,6 and do above procedure, then we will get the desired answer=92399999990760.

 
 My solution:

#include<bits/stdc++.h>
#define PI acos(-1.0)
#define M 346350
#define S(a) scanf("%lld",&a)
#define clr(a) memset(a,0,sizeof(a))
#define pb(x) push_back(x)
#define all(v) v.begin(),v.end()
#define fr(i,a,n) for(i=a;i<=n;i++)
using namespace std;
typedef long long ll;

int main()
{
    ll n,m,t,a,b,i,j,k,x,tc,cs=1,fac[14],fr[12],dig[14];

    fac[0]=1;
    dig[0]=1;
    fr(i,1,12)
    {
        fac[i]=fac[i-1]*i;
        dig[i]=dig[i-1]*10;
    }

    while(S(n) && n)
    {
        vector<ll>v;
        clr(fr);
        fr(i,1,n)
        {
            S(x);
            fr[x]++;
            if(fr[x]==1)v.pb(x);
        }
     
        int sz=v.size();
        ll sm=0;
        fr(i,0,sz-1)
        {
            fr[v[i]]--;
            ll fc=fac[n-1];

            fr(j,0,sz-1)
            {
                fc/=fac[fr[v[j]]];
            }

            ll mul=0;
            fr(j,0,n-1)
            {
                mul+=dig[j];
            }
            fr[v[i]]++;

            sm+=(v[i]*fc*mul);
        }
     printf("%lld\n",sm);
    }
  return 0;
}