Hdu.5211.mutiple (Mathematical derivation && finding all factors of a number in a logn time period)

Mutiple accepts:476 submissions:1025Time limit:4000/2000 MS (java/others) Memory limit:65536/65536 K (java/others)Problem description

WLD has a sequencea[1..  N]    , for each1≤i<n    And he wants you to find out a minimalJ(later with the markF(i)   expression) to meetI<j≤n    MakeaJ Forai The multiple (i.e.aJ MoDai =0), if there is no suchJ, then this makesF(i)   = 0 Guaranteed1≤n≤10000,1≤ai≤10000      For any1≤i≤n    , and for any1≤i,j≤n(i! ) =J)            Meet ai  ! = aj  

Enter a description

Multiple sets of data (up toTenGroup) for each group of data: first row: one numberNRepresents the number of numbers next line: n number, in order a1,a2,... ,an          

Output description

For each set of data:  The sum of output F(i)    (for  1≤i<n )

Input sample

41 3 2 4

Output sample

6

Hint

F (1) =2f (2) =0f (3) =4f (4) =0

1#include <stdio.h>2#include <string.h>3#include <vector>4 Const intM = 1e4 +Ten ;5Std::vector <int>G[m];6 intF[m];7 intA[m];8 intN;9 Ten voidInit () One { A      for(inti =1; I <=10000; i + +) { -          for(intj = i; J <=10000; J + =i) { - G[j].push_back (i); the         } -     } - } -  + intMain () - { +    //freopen ("A.txt", "R", stdin); A init (); at      while(~ scanf ("%d", &N)) { -          for(inti =1; I <= N; i + +) scanf ("%d", &a[i]); -Memset (F,0,sizeof(f)); -         intsum =0 ; -          for(inti = n; i >0; I--) { -Sum + =F[a[i]]; in            //printf ("a[%d]=%d, F =%d\n", I, A[i], f[a[i]); -              for(intj =0; J < G[a[i]].size (); J + +) { tof[G[a[i]][j]]=i; +             } -         } theprintf ("%d\n", sum); *     } $     return 0 ;Panax Notoginseng}

View Code

Jeshen teaches the play, actually is asks 1, 2, ... the factor complexity of each number of n is reduced to O (n/1 + n/2 + n/3 ... n/n) = O (Nlogn), so the average is O (logn).

Hdu.5211.mutiple (Mathematical derivation && finding all factors of a number in a logn time period)