nyoj139--I rank (comtop unfold)

I am in the first few time limits: ms | Memory limit:65535 KB Difficulty:3

Describe

Now there are "ABCDEFGHIJKL" 12 characters, all of which are arranged in a dictionary order, give any sort of arrangement, say this arrangement in all permutations is the number of small?

Input

The

first line has an integer n (0<n<=10000);
Then there are n rows, each line being an arrangement;

Output

outputs an integer m, one row, and M indicates the number of permutations;

Sample input

3abcdefghijklhgebkflacdjigfkedhjblcia

Sample output

1302715242260726926

Source

[Miao Building] Original

Uploaded by

Miao-dong building

Meng B;

#include <cstdio>#include<cstring>inta[ the];voiddeal () {a[1]=1;  for(intI=2; i<= A; i++) A[i]=a[i-1]*i;}intMain () {deal (); intT;SCANF ("%d", &t);  while(t--)    {        Charstr[ the]; scanf ("%s", str);intlen=strlen (str); intREC =0;  for(intI=0; i<len; i++)        {            intCount=0;  for(intj=i+1; j<len; J + +)                if(Str[j]<str[i]) count++; Rec+ = count*a[len-1-i]; } printf ("%d\n", rec+1); }     return 0;}

#include <cstdio>#include<cstring>intjc[ -]={0,1,2,6, -, -,720,5040,40320,362880,3628800,39916800,479001600};intMain () {Charnormal[ -]="ABCDEFGHIJKL"; intT scanf"%d", &t);  while(t--)    {        intv[ -]; memset (V,0,sizeof(v)); Charstr[ -]; scanf"%s", str); intlen=strlen (str); intsum=0;  for(intI=0; i<len; i++) {V[str[i]-'a']=1; if(str[i]-normal[i]==0)            {                Continue; }            Else            {                intD=0;intp=str[i]-'a';  for(intk=p-1; k>=0; k--)                    if(v[k]==0) d++; intj=i+1; intt= A-J; //printf ("%d\n", j+t);sum=sum+d*Jc[t]; }} printf ("%d\n", sum+1); }    return 0;}

falied

nyoj139--I rank (comtop unfold)