UVA

原题：

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 to o slow to ever
pass the time limit. However hard you try, you just can’t discover the prop er break-off conditions that
would bring down the numb er of iterations to within acceptable limits.
Now if the range of input values is not to o big, there is a way out of this. Let your PC rattle for half
an hour and pro duce a table of answers for all p ossible input values, enco de 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 rememb er: In love and programming contests everything is p ermitted).
Faced with this problem during one programming contest, Jimmy decided to apply such a ’technique’. 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 lo oked hop eless, until he discovered the following property regarding the answers: the answers where calculated from two integers, but whenever the two
input values had a common factor, the answer could b e 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) wherexandyare both integers in the range
[1, N]. When he knowsAnswer(x, y), he can easily deriveAnswer(k∗x, k∗y), wherekis any integer
from it by applying some simple calculations involvingAnswer(x, y) andk.
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) andAnswer(4,1). The other 5 can be derived from them (Answer(2,2),Answer(3,3) andAnswer(4,4) fromAnswer(1,1),Answer(2,4) from
Answer(1,2), andAnswer(4,2) fromAnswer(2,1)). Note that the functionAnswer 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 file 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 ofN.
Sample Input
250
Sample Output
3
19

题意：

       对于两个整数x,y，定义二元组(x,y)与二元组(x*k,y*k) （k为正整数）重复（即只保留(x,y)即可），对于1<=x,y<=n，求最简有多少个不重复的二元组。

思路：

        本质为x,y互素的二元组有多少个，设当x<y时，有f(n)个这样的二元组，则可转化为欧拉函数求解小于y且与y互素的正整数有多少个，即f(n)=phi(2)+phi(3)+……+phi(n)。同理x>y时，f(n)=phi(2)+phi(3)+……+phi(n)。当x=y时，只有x=y=1时互素，则答案为2*f(n)+1。

#include <iostream>#include <iomanip>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <deque>#include <string>#include <cmath>#include <vector>#include <utility>#include <set>#include <map>#include <climits>//#pragma comment(linker, "/STACK:1024000000,1024000000")#define INF 2147483647using namespace std;typedef long long ll;int N,phi[50005];void phi_table(int n,ll &ans){    memset(phi,0,sizeof(phi));    phi[1]=1;    for(int i=2; i<=n; i++)    {        if(!phi[i])            for(int j=i; j<=n; j+=i)            {                if(!phi[j])                    phi[j]=j;                phi[j]=phi[j]/i*(i-1);            }        ans+=phi[i];    }}int main(){    while(scanf("%d",&N)&&N)    {        ll res=0;        phi_table(N,res);        printf("%lld\n",2*res+1);    }    return 0;}