Description

Given the value of N, you will have to find the value of G. The meaning of G is given in the following code

G=0;

for(i=1;i<N;i++)

for(j=i+1;j<=N;j++)

G+=gcd(i,j);

/* Here gcd () is a function that finds the greatest common divisor of the two input numbers */

Input

The input file contains at most 20000 lines of inputs. each line contains an integer N (1 <n <1000001 ). the meaning of N is given in the problem statement. input is terminated by a line containing a single zero.

Output

For each line of input produce one line of output. This line contains the value of G for the corresponding N. The value of G will fit in a 64-bit signed integer.

Sample Input

10 100 200000 0

Sample output

67 13015 143295493160

Source

Contest for 2010 lecture II

Question: Give the number N, and calculate the sum of all the number pairs that satisfy 1 <= I <j <= n, and the gcd (I, j) corresponding to (I, j.

Idea: set F (n) = gcd (1, N) + gcd (2, n) + gcd (3, n) +... + Gcd (n-1, n ). Then the answer S (n) = F (2) + f (3) + f (4) +... + F (n ).

Note that the value of all gcd (x, n) is an approximate number of N, so we can classify them according to this approximate number. G (n, I) is used to represent the number of positive integers x that meet gcd (x, n) = I and X <n. Then f (n) = {I × G (n, I) | I is the approximate number of n }. Note that G (n, I) = PHI (n, I ).

If I is enumerated for each n, it will definitely time out according to the data range. We should enumerate I first, and then find the multiple n of I. In this way, the time complexity will be further reduced.

1/* 2 * Author: Joshua 3 * created time: October 16, Sunday, September 07, 2014 4 * File Name: fzu1969.cpp 5 */6 # include <cstdio> 7 # include <cstring> 8 typedef long ll; 9 # define maxn 100000510 int Phi [maxn], a [maxn], f [maxn]; 11 ll s [maxn]; 12 bool P [maxn]; 13 int tot = 0; 14 15 void PRI () 16 {17 memset (P, 1, sizeof (p); 18 for (INT I = 2; I <maxn; ++ I) 19 if (P [I]) 20 {21 for (Int J = I <1; j <maxn; j + = I) 22 p [J] = false; 23} 24 for (INT I = 2; I <maxn; ++ I) 25 if (P [I]) A [++ tot] = I; 26} 27 28 void phi_table () 29 {30 for (INT I = 2; I <maxn; ++ I) 31 {32 Phi [I] = I; 33 int temp = I; 34 For (Int J = 1; j <= tot; ++ J) 35 {36 IF (temp = 1) break; 37 If (P [temp]) 38 {39 Phi [I]/= temp; 40 Phi [I] * = temp-1; 41 break; 42} 43 If (TEMP % A [J] = 0) 44 {45 Phi [I]/= A [J]; 46 Phi [I] * = A [J]-1; 47 While (TEMP % A [J] = 0) temp/= A [J]; 48} 49} 50} 51} 52 53 void solve () 54 {55 PRI (); 56 phi_table (); 57 for (INT I = 1; I <maxn; ++ I) 58 for (Int J = I; j <maxn; J + = I) 59 f [J] + = I * Phi [J/I]; 60 for (INT I = 1; I <maxn; ++ I) 61 s [I] + = s [I-1] + F [I]; 62} 63 64 int main () 65 {66 int N; 67 solve (); 68 while (scanf ("% d", & N) 69 printf ("% LLD \ n", s [N]); 70 return 0; 71}

Fzu1969 GCD extreme is similar to uva10561