uva10820 Send a Table

When participating in programming contests, you sometimes face the
following problem: You know how to calcutale the output for the given
input values, but your algorithm is way too slow to ever pass the time
limit. However hard you try, you just can’t discover the proper
break-off conditions that would bring down the number of iterations to
within acceptable limits. Now if the range of input values is not too
big, there is a way out of this. Let your PC rattle for half an hour
and produce a table of answers for all possible input values, encode
this table into a program, submit it to the judge, et voila: Accepted
in 0.000 seconds! (Some would argue that this is cheating, but
remember: In love and programming contests everything is permitted).
Faced with this problem during one programming contest, Jimmy decided
to apply such a ‘tech- nique’. But however hard he tried, he wasn’t
able to squeeze all his pre-calculated values into a program small
enough to pass the judge. The situation looked hopeless, until he
discovered the following prop- erty regarding the answers: the answers
where calculated from two integers, but whenever the two input values
had a common factor, the answer could be easily derived from the
answer for which the input values were divided by that factor. To put
it in other words: Say Jimmy had to calculate a function Answer ( x;y
) where x and y are both integers in the range [1 ;N ]. When he knows
Answer ( x;y ), he can easily derive Answer ( k  x;k  y ), where k
is any integer from it by applying some simple calculations involving
Answer ( x;y ) and k . For example if N
= 4, he only needs to know the answers for 11 out of the 16 possible input value combinations: Answer (1 ; 1), Answer (1 ; 2), Answer (2 ;
1), Answer (1 ; 3), Answer (2 ; 3), Answer (3 ; 2), Answer (3 ; 1),
Answer (1 ; 4), Answer (3 ; 4), Answer (4 ; 3) and Answer (4 ; 1). The
other 5 can be de- rived from them ( Answer (2 ; 2), Answer (3 ; 3)
and Answer (4 ; 4) from Answer (1 ; 1), Answer (2 ; 4) from Answer (1
; 2), and Answer (4 ; 2) from Answer (2 ; 1)). Note that the function
Answer is not symmetric, so Answer (3 ; 2) can not be derived from
Answer (2 ; 3). Now what we want you to do is: for any values of N
from 1 upto and including 50000, give the number of function Jimmy has
to pre-calculate. Input The input le contains at most 600 lines of
inputs. Each line contains an integer less than 50001 which indicates
the value of N . Input is terminated by a line which contains a zero.
This line should not be processed. Output For each line of input
produce one line of output. This line contains an integer which
indicates how many values Jimmy has to pre-calculate for a certain
value of N .

首先很容易看出有且只有x,y互质才需要加入答案，所以本质是求互质的数对(x,y)的个数。
不妨设x< y，对于某个y，答案的个数是小于他且与他互质的数，即phi(y)。
所以答案为2*Σphi(2..n)+1
最后的加上的1是(1,1)。

#include<cstdio>#include<cstring>const int maxn=50005;long long phi[50010],ans[50010];int main(){    int i,j,k,m,n,p,q,x,y,z;    phi[1]=1;    for (i=2;i<=maxn;i++)      if (!phi[i])        for (j=i;j<=maxn;j+=i)        {            if (!phi[j]) phi[j]=j;            phi[j]=phi[j]/i*(i-1);        }    for (i=2;i<=maxn;i++)      ans[i]=ans[i-1]+phi[i];    while (scanf("%d",&n)&&n)      printf("%lld\n",ans[n]*2+1);}