UVa 10820 (play table, Euler function) Send a table

Test instructions

The background of the topic is omitted, it is easy to translate the problem into, give N, n the number of ordered pairs (x, y) of the logarithm of the mutual pigment.

Analysis:

Problems can also continue to be transformed.

According to symmetry, we can assume that x<y, when x=y, satisfies only the conditions (1, 1).

Set F (n) as the set s{(x, y) | x<y and X, Y for the number of reciprocal, then the answer is 2f (n) +1

An F (n) expression is:

, where φ (n) is the Euler function

Here are some introduction to Euler's functions

1#include <cstdio>2 3 Const intMAXN =50000;4 5 intPHI[MAXN +Ten], SUM[MAXN +Ten];6 7 voidPhi_table (intN)8 {9phi[1] =0;Ten      for(inti =2; I <= MAXN; ++i)if(!phi[i])//description I is prime number One     { A          for(intj = i; J <= Maxn; J + =i) -         { -             if(!phi[j]) phi[j] =J; thePHI[J] = phi[j]/I * (i-1); -         } -     } - } +  - intMain () + { A phi_table (MAXN); atsum[1] = phi[1]; -      for(inti =2; I <= MAXN; ++i) Sum[i] = sum[i-1] +Phi[i]; -      -     intN; -      while(SCANF ("%d", &n) = =1&& N) printf ("%d\n",2*sum[n]+1); -      in     return 0; -}

code June

UVa 10820 (play table, Euler function) Send a table